what is the prediction equation for the median income and incarcerat.pdf
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what is the prediction equation for the median income and incarceration rate...and the prediction equation for the total crime and incarceration rate? Incarceration Rate Median Violent Crime Prop crime Total crime per 100,00 Adults Income Rate Rate Rate 2013 2010-2013 2012 2012 2012 State Alabama 910 43330 449.9 3502.2 3952.1 Arkansas 1010 40877 469.1 3660.1 4129.2 Connecticut 620 67807 283 2140 2423 Florida 990 47106 487.1 3276.7 3763.8 Idaho 860 49952 207.9 1983.5 2191.4 Iowa 530 53364 263.9 2271.8 2535.7 Louisiana 1420 40844 496.9 3540.6 4037.5 Massachusetts 400 64555 405.5 2153 2558.5 Mississippi 950 40338 260.8 2811 3071.8 Nebraska 600 55107 259.4 2754.9 3014.3 New Jersey 540 65321 290.2 2047.3 2337.5 North Carolina 730 44254 353.4 3369.5 3722.9 Oklahoma 740 47282 469.3 3401 3870.3 Rhode Island 400 55158 252.4 2572.3 2824.7 Tennessee 960 42785 643.6 3371.4 4015 Vermont 410 56175 142.6 2398.7 2541.3 West Virginia 660 43361 316.3 2364.9 2681.2 Mean 748.82 50448.00 355.96 2801.11 3157.07 SD 273.52 9056.11 130.62 601.63 703.41 Incarceration Rate Median Violent Crime Prop crime Total crime per 100,00 Adults Income Rate Rate Rate 2013 2010-2013 2012 2012 2012 State Alabama 910 43330 449.9 3502.2 3952.1 Arkansas 1010 40877 469.1 3660.1 4129.2 Connecticut 620 67807 283 2140 2423 Florida 990 47106 487.1 3276.7 3763.8 Idaho 860 49952 207.9 1983.5 2191.4 Iowa 530 53364 263.9 2271.8 2535.7 Louisiana 1420 40844 496.9 3540.6 4037.5 Massachusetts 400 64555 405.5 2153 2558.5 Mississippi 950 40338 260.8 2811 3071.8 Nebraska 600 55107 259.4 2754.9 3014.3 New Jersey 540 65321 290.2 2047.3 2337.5 North Carolina 730 44254 353.4 3369.5 3722.9 Oklahoma 740 47282 469.3 3401 3870.3 Rhode Island 400 55158 252.4 2572.3 2824.7 Tennessee 960 42785 643.6 3371.4 4015 Vermont 410 56175 142.6 2398.7 2541.3 West Virginia 660 43361 316.3 2364.9 2681.2 Mean 748.82 50448.00 355.96 2801.11 3157.07 SD 273.52 9056.11 130.62 601.63 703.41 Solution what is the prediction equation for the median income and incarceration rate...and the prediction equation for the total crime and incarceration rate? Regression Analysis r² 0.536 n 17 r -0.732 k 1 Std. Error 192.463 Dep. Var. Incarceration Rate ANOVA table Source SS df MS F p-value Regression 641,346.1758 1 641,346.1758 17.31 .0008 Residual 555,630.2948 15 37,042.0197 Total 1,196,976.4706 16 Regression output confidence interval variables coefficients std. error t (df=15) p-value 95% lower 95% upper Intercept 1,864.1151 272.0681 6.852 5.49E-06 1,284.2157 2,444.0145 Median income -0.0221 0.0053 -4.161 .0008 -0.0334 -0.0108 Incarceration Rate =1,864.1151 - 0.0221*median income Regression Analysis r² 0.491 n 17 r 0.701 k 1 Std. Error 201.486 Dep. Var. Incarceration Rate ANOVA table Source SS df MS F p-value Regression 588,028.9910 1 588,028.9910 14.48 .0017 Residual 608,947.4796 15 40,596.4986 Total 1,196,976.4706 16 Regression output confidence interval variables coefficients std. error t (df=15) p-value 95% lower 95% upper Intercept -.
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2015 HIV SURVEILLANCE DATA IN THE U.S. In.docx
2015 HIV
SURVEILLANCE
DATA
IN THE U.S.
In 2014, an estimated 44,073 people were diagnosed with HIV infection in the
United States. 1 in 8 individuals living with HIV don’t know their status. (cdc.gov,
June 2016)
IN TEXAS
As of 2015, 82,745 individuals were known to be living with HIV in Texas.
In 2015, 4,486 cases of HIV were newly diagnosed in Texas.
DISPARITIES
Gay and Bisexual men continue to be most affected by the HIV epidemic in the
U.S.
At current rates.
• 1 in 6 MSM will be diagnosed with HIV in their lifetime,
• 1 in 2 black MSM
• 1 in 4 Latino MSM
• 1 in 11 white MSM (Center for Disease Control and Prevention 2016)
HIV DIAGNOSIS RATES BY SEX,
RACE/ETHNICITY AND AGE
GROUP, 2004-2015
2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015
Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate*
Sex
Female 962 8.4 905 7.7 980 8.2 939 8 965 8 992 8 888 7 840 6 816 6 823 6 842 6.1
Male 3,368 29.8 3,071 26.5 3,139 26.6 3,239 27 3,405 28 3,485 28 3,422 27 3,520 27 3,541 27 3,582 27 3644 26.7
Total 4,330 3,976 4,119 4,178 4,370 4,477 4,310 4,360 4,357 4,405 4,86
Race
White 1,190 10.6 995 8.8 961 8.4 976 9 944 8 923 8 879 8 909 8 935 8 919 8 940 7.8
Black 1,598 60.7 1,604 57.9 1,616 57.4 1,696 59 1,735 59 1,764 59 1,595 52 1,595 51 1,634 51 1,649 51 1657 49.8
Hispanic 1,366 17.1 1,233 14.8 1,347 15.6 1,333 15 1,475 16 1,555 16 1,585 16 1,638 16 1,581 16 1,650 16 1706 16
Other 36 4.1 35 3.8 42 4.3 47 5 44 4 58 5 73 6 64 5 78 6 83 6 94 6.5
Unknown 140 0.0 109 0.0 153 0.0 126 0 172 0 177 0 178 0 154 0 129 0 104 0 89
Age Group
0 - 9 12 0.3 12 0.3 23 0.6 17 0.45 19 0.5 14 0.36 24 0.62 16 0.41 14 0.36 12 0.3 13 0.3
10 - 14 5 0.3 8 0.5 5 0.3 2 0.11 6 0.32 5 0.27 5 0.26 9 0.46 6 0.31 7 0.35 4 .02
15 - 19 143 8.2 172 9.6 186 10.2 197 10.67 254 13.55 252 13.38 230 12.23 242 12.86 206 10.89 215 11.29 221 11.4
20 - 24 491 28.1 495 28.1 563 31.9 664 37.35 724 40.27 797 43.86 800 42.64 812 41.94 823 41.71 917 45.84 876 43.4
25 - 29 662 40.3 618 36.3 657 37.4 675 37.29 691 37.52 676 36.48 692 36.85 734 38.6 810 42.13 847 43.04 851 42.1
30 - 34 665 40.2 580 35.0 571 34.2 577 34.23 628 36.42 644 36.58 564 31 632 33.8 639 33.32 594 30.34 691 34.7
35 - 39 735 45.3 602 36.0 619 36.0 592 33.86 549 31.13 532 30.17 494 28.22 466 26.5 450 25.34 488 26.96 503 27.0
40 - 44 649 37.9 583 34.2 554 32.8 512 30.53 518 30.86 509 30.03 464 26.53 458 25.62 417 23.04 404 22.14 366 20
45+ 968 13.1 906 11.8 941 11.8 942 11.49 981 11.61 1048 12.14 1037 11.65 991 10.88 992 10.65 921 9.66 961 9.8
Unknown 0 0.0 0 0.0 0 0.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Total 4330 19.0 3976 17.0 4119 17.3 4178 17.19 4370 17.62 4477 17.8 4310 16.8 ...
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2015 HIV SURVEILLANCE DATA IN THE U.S. In.docx
2015 HIV
SURVEILLANCE
DATA
IN THE U.S.
In 2014, an estimated 44,073 people were diagnosed with HIV infection in the
United States. 1 in 8 individuals living with HIV don’t know their status. (cdc.gov,
June 2016)
IN TEXAS
As of 2015, 82,745 individuals were known to be living with HIV in Texas.
In 2015, 4,486 cases of HIV were newly diagnosed in Texas.
DISPARITIES
Gay and Bisexual men continue to be most affected by the HIV epidemic in the
U.S.
At current rates.
• 1 in 6 MSM will be diagnosed with HIV in their lifetime,
• 1 in 2 black MSM
• 1 in 4 Latino MSM
• 1 in 11 white MSM (Center for Disease Control and Prevention 2016)
HIV DIAGNOSIS RATES BY SEX,
RACE/ETHNICITY AND AGE
GROUP, 2004-2015
2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015
Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate* Cases Rate*
Sex
Female 962 8.4 905 7.7 980 8.2 939 8 965 8 992 8 888 7 840 6 816 6 823 6 842 6.1
Male 3,368 29.8 3,071 26.5 3,139 26.6 3,239 27 3,405 28 3,485 28 3,422 27 3,520 27 3,541 27 3,582 27 3644 26.7
Total 4,330 3,976 4,119 4,178 4,370 4,477 4,310 4,360 4,357 4,405 4,86
Race
White 1,190 10.6 995 8.8 961 8.4 976 9 944 8 923 8 879 8 909 8 935 8 919 8 940 7.8
Black 1,598 60.7 1,604 57.9 1,616 57.4 1,696 59 1,735 59 1,764 59 1,595 52 1,595 51 1,634 51 1,649 51 1657 49.8
Hispanic 1,366 17.1 1,233 14.8 1,347 15.6 1,333 15 1,475 16 1,555 16 1,585 16 1,638 16 1,581 16 1,650 16 1706 16
Other 36 4.1 35 3.8 42 4.3 47 5 44 4 58 5 73 6 64 5 78 6 83 6 94 6.5
Unknown 140 0.0 109 0.0 153 0.0 126 0 172 0 177 0 178 0 154 0 129 0 104 0 89
Age Group
0 - 9 12 0.3 12 0.3 23 0.6 17 0.45 19 0.5 14 0.36 24 0.62 16 0.41 14 0.36 12 0.3 13 0.3
10 - 14 5 0.3 8 0.5 5 0.3 2 0.11 6 0.32 5 0.27 5 0.26 9 0.46 6 0.31 7 0.35 4 .02
15 - 19 143 8.2 172 9.6 186 10.2 197 10.67 254 13.55 252 13.38 230 12.23 242 12.86 206 10.89 215 11.29 221 11.4
20 - 24 491 28.1 495 28.1 563 31.9 664 37.35 724 40.27 797 43.86 800 42.64 812 41.94 823 41.71 917 45.84 876 43.4
25 - 29 662 40.3 618 36.3 657 37.4 675 37.29 691 37.52 676 36.48 692 36.85 734 38.6 810 42.13 847 43.04 851 42.1
30 - 34 665 40.2 580 35.0 571 34.2 577 34.23 628 36.42 644 36.58 564 31 632 33.8 639 33.32 594 30.34 691 34.7
35 - 39 735 45.3 602 36.0 619 36.0 592 33.86 549 31.13 532 30.17 494 28.22 466 26.5 450 25.34 488 26.96 503 27.0
40 - 44 649 37.9 583 34.2 554 32.8 512 30.53 518 30.86 509 30.03 464 26.53 458 25.62 417 23.04 404 22.14 366 20
45+ 968 13.1 906 11.8 941 11.8 942 11.49 981 11.61 1048 12.14 1037 11.65 991 10.88 992 10.65 921 9.66 961 9.8
Unknown 0 0.0 0 0.0 0 0.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Total 4330 19.0 3976 17.0 4119 17.3 4178 17.19 4370 17.62 4477 17.8 4310 16.8 ...
According to the IRS, 1.1 of tax returns will be audited in 2011. A.pdf
According to the IRS, 1.1% of tax returns will be audited in 2011. A random sample of 40 tax
returns was selected. Use the Poisson distribution to approximate the binomial distribution to
determine the probability that more than two of these tax returns will be audited.0.1215
Solution
for poisson distribution, mean = np = 40(0.011) = 0.44
by using excel function =1-POISSON(1,0.44,1), P(x>=2) = 0.0726.
Write a SELECT statement that returns a single value that represents.pdf
Write a SELECT statement that returns a single value that represents the sum of the largest
unpaid invoices submitted by each vendor. Use a common table expression(cte) that returns
MAX(InvoiceTotal) grouped by VendorID, filtering for invoices with a balanced due.
Solution
WITH INVOICE_CTE (VendorID,maxInvoiceTotal)
AS
(
Select VendorID,MAX(InvoiceTotal) as maxInvoiceTotal
from Invoice
where due>0
)
Select MAX(maxInvoiceTotal) from
INVOICE_CTE.
which represents the ground state for the N- ion i. w.ia ere sm.pdf
which represents the ground state for the N- ion
i. w.ia ere sms he Morral su rrie ...dae ... lilili
Solution
The ground state for the N- ion
option e..
The ground state of an atom or ion is the state in which all its electrons are found in the possible
lowest energy orbitals .
your rest of the image of questions are not clear ..
What is the ecological and environmental importance of salinityoxyg.pdf
What is the ecological and environmental importance of salinity/oxygen consumption of grass
shrimp?
Solution
Grass shrimp acts as detritivores,grass shrimp aid in the mechanical break down of organic
matter such as fibrous plant materials into tiny particles that are suspended in the water column,
and thus provides rich food source for smaller organisms. Predation : In estuaries, numerous fish
species and other aquatic carnivores , eat large quantities of grass shrimp. Grass shrimp are hosts
for numerous species of parasites and ectocommensals. Grass shrimp is pollution tolerant and
live in areas of high salinity. It consumes algae and seagrasses, therefore plays an important role
in the ecology of the estuary..
WEP has vulnerabilities. Which of the following is not a reason why .pdf
WEP has vulnerabilities. Which of the following is not a reason why it is vulnerable?
A. Shared WEP keys among all clients
B. An RC4 engine not properly initialized
C. 20-bit initialization vector
D. 40-bit WEP keys
Solution
C. One issue with WEP is the initialization vector (IV), it was 24 bits, not 20. Answers A, B, and
D detail some of the vulnerabilities of WEP. For example, WEP uses a single shared key among
all clients, which means that you are authenticating groups, not devices or single users. Also,
RC4 is the correct encryption type and can be implemented in 40- or 104-bit configuration, but
WEP does not properly initialize it. This means that the key values roll over and are predictable.
Finally, a 24-bit IV vector is too short, and a 40-bit key is weak..
Two astronauts are 2.40 m apart in their spaceship. One speaks to the.pdf
Two astronauts are 2.40 m apart in their spaceship. One speaks to the other. The conversation is
transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at 343
m/s through the air between the astronauts equals the time it takes for the electromagnetic waves
to travel to the earth. How far away from the earth is the spaceship Number Units
Solution
time taken by sound to travel 2.40 m,
t = 2.40 / 343 s
spaceship is at the distance at light can travel in time t.
distance = speed x time
= (3 x 10^8 ) (2.40 / 343)
= 2.099 x 10^6 m.
The ratio of the probability of disease in an exposed group to the p.pdf
The ratio of the probability of disease in an exposed group to the probability of disease in an
unexposed group is called the: 1. Relative risk 2. Odds ratio 3. Risk difference 4. Attributable
risk
Solution
The ratio of the probability of disease in an exposed group to the probability of disease in an
unexposed group is called relative risk.
RR=P(diseased/exposed)/P(diseased/unexposed).
The output should now look like this Row 1 sum 30 Row 2 sum.pdf
The output should now look like this:
Row 1 sum: 30
Row 2 sum: 48
Row 3 sum: 55
Count: 12
Grand total: 133
Mean: 11.08
Even numbers: 7
Odd numbers: 5
How to add even and odd numbers counts to existing code
public class Problem {
public static void main(String [] args)
{
int iVal;
int iRow;
int iRowSum;
int iToken;
int totalSum = 0;
int totalEle = 0;
iRowSum = 0;
iRow = 1;
try
{
Scanner ifsInput = new Scanner(new File(\"c:\\home\\student\\lastname/9_1_Input.txt\"));
while(ifsInput.hasNextLine())
{
while(ifsInput.hasNextInt())
{
for(iToken = 0; iToken <= 3; iToken++)
{
iVal = ifsInput.nextInt();
iRowSum = iRowSum + iVal;
// increasing count of total element
totalEle++;
}
// adding into totalSum
totalSum = totalSum + iRowSum;
System.out.println(\"Row \" + iRow + \" sum: \" + iRowSum);
iRowSum = 0;
iRow++;
}
}
// added new code to print grand total and mean value
System.out.println(\"Count: \"+totalEle);
System.out.println(\"Grand total: \"+totalSum);
System.out.println(\"Mean: \"+String.format(\".2f\", (double)/totalSum/totalEle);
}
catch (FileNotFoundException sMsg)
{
System.out.println(\"The file cannot be found or opened.\");
}
}
}
Solution
Input999.txt( Save this file under D Drive Then the path of the file pointing to it is
D:\\\\Input999.txt)
11 22 33 44
22 33 44 55
33 44 55 66
___________________________
Problem.java
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Problem {
public static void main(String [] args)
{
//Declaring variables
int iVal;
int iRow;
int iRowSum;
int iToken;
double mean=0.0;
int evenCount=0,oddCount=0;
int totalSum = 0;
int totalEle = 0;
iRowSum = 0;
iRow = 1;
try
{
Scanner ifsInput = new Scanner(new File(\"D:\\\\Input999.txt\"));
while(ifsInput.hasNextLine())
{
while(ifsInput.hasNextInt())
{
for(iToken = 0; iToken <= 3; iToken++)
{
iVal = ifsInput.nextInt();
//Calculating the even count
if(iVal%2==0)
evenCount++;
//Calculating the odd count
else
oddCount++;
//Calculating the sum of elements in each row
iRowSum = iRowSum + iVal;
// increasing count of total element
totalEle++;
}
// adding into totalSum
totalSum = totalSum + iRowSum;
//Displaying the sum of elements in each row
System.out.println(\"Row \" + iRow + \" sum: \" + iRowSum);
iRowSum = 0;
iRow++;
}
}
//Displaying the number of elements
System.out.println(\"Count: \"+totalEle);
//Displaying the sum of all elements
System.out.println(\"Grand total: \"+totalSum);
//Calculating the mean
mean=(double)totalSum/totalEle;
//Displaying the mean
System.out.printf(\"Mean: %.2f\ \",mean);
//Displaying the number of even nos
System.out.println(\"Even Numbers :\"+evenCount);
//Displaying the number of odd nos
System.out.println(\"Odd Numbers :\"+oddCount);
}
catch (FileNotFoundException sMsg)
{
System.out.println(\"The file cannot be found or opened.\");
}
}
}
_________________________________
Output:
Row 1 sum: 110
Row 2 sum: 154
Row 3 sum: 198
Count: 12
Grand total: 462
Mean: 38.50
Even Numbers :6
Odd Numbers :6
______.
The end problem in eukaryotic DNA replicationa. is solved by t.pdf
The \"end problem\" in eukaryotic DNA replication
a. is solved by telomerase
b. refers to loss of chromosome ends during replication
c. is solved by an RNA dependent DNA polymerase
d. A and B
e. all of the above
Solution
Answer : The correct answer is E ( All of the above)
Explanation :
Telomerases are the ribonucleoprotein complexes containing a small RNA that serves as a
template for addition of new six nucleotide repeats. Or Telomerase, the enzyme that synthesizes
the telomere ends of linear chromosomes, is a specialized reverse transcriptase that contains an
internal RNA template. Telomeres undergo cycles of shortening of the lagging strands due to the
inability to complete synthesis.
Most normal cells lose telomerase activity with increaseing age of the organism and repeated
generations of the cell division. In other words, after certain number of divisions normal cells
reaches the Hayflicks limit beyond which cells will no longer devide instead they die.
In germ-line cells, which contain telomerase activity, telomere lengths are maintained; for
example, telomerase is highly expressed only in cells that need to divide regularly especially in
male sperm cells. Whereas somatic cells lack telomerase activity due to which they die after
certain number of generations.
In cancer cells/tumor cells, end replication problems observed in normal cells will be taken care
by the telomerase enzyme which makes the cells to bypass the Hayflick limit and become
immortal, thus avoiding cell death Many cancer cells are considered \'immortal\' because
telomerase activity supports them to live longer than somatic cells. 90% of the cancer cells retain
the ability to express telomerase and acquires the ability to proloferate indefinitely..
Suppose that one obtained the following DNA sequence data for the fou.pdf
Suppose that one obtained the following DNA sequence data for the four fly families and a non-
fly insect (the outgroup) from a gene that is not related to morphology. We will call the DNA
sequence positions 1 through 10, for reference (in fly family A, position 1 is a T, while position
10 is an A). Which of the characters (nucleotide positions) are variable among the fly families
(not counting the outgroup? Which of the characters (nucleotide positions) are phylogenetically
informative for cladistics? (Phylogenetically informative traits are those for which there is
variation in the group of interest, and for which there is evidence of synapomorphies.)
Solution
Purines are 2 ringed structure= A,G whereas, pyramidines aresingle ringed structure = C,T
Generally A = T (A double bonds with T) and G=C (G triple H bonds with C).
Mutation causes a change in theDNA sequence. Two types
1. Transversions: Purine instead of a pyramidine or vice versa (Less frequent so more important
for cladistics) A/G to C/T or vice versa. 2 ringed structure to 1 ringed structure
See this is seen in all fly families at 7th(ABCD:AGGC) , 8th (ABCD:CTAC) and 9th
(ABCD:GCTG)positions
2. Transitions: Purine instead of a puirne and pyramidine instead of a pyramidine (Easy and
frequent)
The Nucleotide positions are Bolded in the table are variable among the fly families. Among all
Fly family C is most variable. (Take A, B, C, D and leave Out group).
Phylogenetically informative traits are those for which there is variation in the group of interest
and for which there is evidence of synapomorphies. All this is to know the members of a group
which share a common history and to know their closely related species. Synapomorphy means
any characteristic which is present in an ancestral species and that is shared exclusively (either in
more or less modified form) by its evolutionary descendants.
Mostly there is a much variation in the 7th, 8th and 9th position of the nucleotide in all fly
families sequence so this will be more phylogenetically informative for cladistics and the 5th
sequence is the most stable one. From the sequence fly family C is much more related to the out
group sequence than all familiesFLY FAMILYDNA Sequence alignedpositions
order12345678910ATGCGTTACGABTGCATTGTCACTCCGTAGATTDGCAGTTCCGAOU
T GROUPGCAGTTGATT.
Set up a JavaFX GUI-based program that shows a 10 times 10 grid of la.pdf
Set up a JavaFX GUI-based program that shows a 10 times 10 grid of labels that forms a
multiplication table, with the labels displaying the multiplication problems, rather than the
answers. Provide a text input field and a button with an event handler that reads an integer value
from the text input and changes the CSS styling in some obvious way for all problems in the
table with the given answer. When a new answer is entered and the button is clicked, change all
the labels back to the original style, then change the labels showing problems for the new answer
to the new style. Use CSS, not setters, for all the styling. The main point of this lab is to learn
JavaFX, so the grading will include a heavy emphasis on the neatness and general appearance of
your GUI.
Solution
Code:
//multiTable.java
// import required packages
import java.util.List;
import javafx.application.Application;
import javafx.event.Event;
import javafx.event.EventHandler;
import javafx.geometry.Insets;
import javafx.geometry.Pos;
import javafx.scene.Node;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.control.TextField;
import javafx.scene.input.MouseEvent;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.HBox;
import javafx.stage.Stage;
// class for revrse multiplication table
public class multiTable extends Application
{
// over ride nethod for setup the grid
@Override
public void start(Stage primaryStage)
{
BorderPane pane1 = new BorderPane();
pane1.setTop(getHbox1());
HBox h_box = new HBox(15);
h_box.setPadding(new Insets(15, 15, 15, 15));
h_box.setAlignment(Pos.TOP_CENTER);
h_box.getStyleClass().add(\"hbox2\");
Label labl = new Label(\"Enter Answer: \");
h_box.getChildren().add(labl);
TextField textfield1 = new TextField();
h_box.getChildren().add(textfield1);
GridPane grid_pane1 = setUpGrid();
gridmaker griddp = new gridmaker(grid_pane1);
Button Answer = new Button(\"Find problems\");
Answer.addEventHandler(MouseEvent.MOUSE_CLICKED, new EventHandler()
{
@Override
public void handle(Event arg0)
{
List fact = factors(textfield1);
for (int[] RCx1 : fact) {
Node node = griddp.get_chldren()[RCx1[0]][RCx1[1]];
node.setStyle(\"-fx-background-color: green\");
}
}
});
h_box.getChildren().add(Answer);
pane1.setCenter(h_box);
pane1.setBottom(grid_pane1);
Scene scene = new Scene(pane1, 550, 650);
scene.getStylesheets().add(getClass().getResource(\"application.css\").toExternalForm());
primaryStage.setTitle(\"lab 6\");
primaryStage.setScene(scene);
primaryStage.show();
}
// method for grid layout
private HBox getHbox1()
{
HBox hbox2 = new HBox(15);
hbox2.setPadding(new Insets(15, 15, 15, 15));
hbox2.setAlignment(Pos.TOP_CENTER);
hbox2.getStyleClass().add(\"hbox1\");
Label labl = new Label(\"Multiplication Table\");
hbox2.getChildren().add(labl);
return hbox2;
}
// method to setup a grid
public GridPane setUpGrid() {
GridPane pane1 = new GridPane();
Label[][].
Sample of DNA isolated from bacterium X contains 17.5 of adenine.pdf
Sample of DNA isolated from bacterium X contains 17.5% of adenine bases. What percentages
of guanine, thymine, and cytosine bases would you expect to find in the same sample of DNA?
Match the following types of transport across a biological membrane (designated I, II, or III)
with the statements designated A-D: Simple diffusion (SD) Facilitated diffusion (FD) Active
transport (AT) Write down your answers in the blank space next to each statement below. Note,
that more than one answer may apply each statement: Process is energy dependent: ___ Process
needs transmembrane transporter/channel: ___ Process occurs down the concentration gradient:
___ Processes can establish a concentration gradient: ___
Solution
Answer 2. Given that adenine bases in the isolated DNA of the bacterium is 17.5%. Since
Adenine is known to pair with thymine. So, the percentage of thymine will also be 17.5%.
Adenine and thymine will make 35 % of the total nitrogen bases of the DNA. Rest of the DNA
will be made from Guanine and Cytosine. Guanine and Cytosine are known to pair with each
other. So, their content in the DNA will also be same. 100-35% will give us 65 %. So Guanine
will be 32.5% and Cytosine will be 32.5%
Answer 3.
Process occur down the concentration gradient - Simple Diffusion(SD)
Process need transmembrane transporter/Channel- Facilitated Diffusion (FD)
Process is energy dependent- Active transport (AT)
Process can establish a concentration gradient- Facilitated Diffusion (FD) and Active transport
(AT).
Practice using layouts Anderson, Franceschi import javax..pdf
/* Practice using layouts
Anderson, Franceschi
*/
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
public class NestedLayoutPractice extends JFrame
{
private Container contents;
private Game game;
private BorderLayout bl;
private JLabel bottom;
// ***** Task 1: declare a JPanel named top
// also declare three JButton instance variables
// that will be added to the JPanel top
// these buttons will determine the grid size of the game:
// 3-by-3, 4-by-4, or 5-by-5
// Part 1 student code starts here:
private final JPanel top;
private JButton [] topButtones;
String [] topGridSize =
{\"3-by-3\",
\"4-by-4\",
\"5-by-5\",
};
private JButton reset;
// Part 1 student code ends here.
public NestedLayoutPractice()
{
super(\"Practicing layout managers\");
contents = getContentPane();
// ***** Task 2:
// instantiate the BorderLayout manager bl
// Part 2 student code starts here:
// set the layout manager of the content pane contents to bl:
// bl = new BorderLayout (); (not suer if this is right)
//set.BorderLayout(bl); (not suer if this is right)
game = new Game(3); // instantiating the GamePanel object
// add panel (game) to the center of the content pane
// Part 2 student code ends here.
bottom = new JLabel(\"Have fun playing this Tile Puzzle game\",
SwingConstants.CENTER);
// ***** Task 3:
// instantiate the JPanel component named top
// Part 3 student code starts here:
//top = new JPanle ();(not suer if this is right)
// set the layout of top to a 1-by-3 grid(not suer if this is right)
// panel.setLayout (new GridLayout (3, 1));(not suer if this is right)
// instantiate the JButtons that determine the grid size
// topButtones [0] = new JButton (\"3-by-3 \"); (not suer if this is right)
// add the buttons to JPanel top
//top.add (topButtones[0]);(not suer if this is right)
//top.add (topButtones[1]);
//top.add (topButtones[2]);
// add JPanel top to the content pane as its north component
c.add(top, \"north\");
// Part 3 student code ends here.
// ***** Task 5:
// Note: search for and complete Task 4 before performing this task
// Part 5 student code starts here:
// declare and instantiate an ActionListener
//topButtones[0].addActionListnere(this);(not suer if this is right)
//topButtones[1].addActionListnere(this);
//topButtones[2].addActionListnere(this);
// register the listener on the 3 buttons
// that you declared in Task 1
// Part 5 student code ends here.
contents.add(bottom, BorderLayout.SOUTH);
setSize(325, 325);
setVisible(true);
}
// ***** Task 4:
// create a private inner class that implements ActionListener
// your method should identify which of the 3 buttons
// was the source of the event
// depending on which button was pressed,
// call the setUpGame method of the Game class
// with arguments 3, 4, or 5
// the API of that method is:
// public void setUpGame(int nSides)
// At the end of the method call validate()
// Part 4 student code starts here:
public void actionPreformed(ActionEvent ae)
{
if (ae.getActionCommand(). equ.
Please, I need a correct answer and clear explanation. open image an.pdf
Please, I need a correct answer and clear explanation. open image and zoom for a clear view.
Thanks,
Using the West Jet Financial statements, answer the following questions R1 2-4 p.106-109)
b. Why do you think West Jet does not report a cost of goods sold
c. Explain what “Advanced ticket sales” in the current liabilities section represents
e. How much did West Jet use in 2013 to acquire new aircraft? Which statement did you find this
information on?
RI2-4
(Determination of items from a Canadian company\'s financial statements)
Calgary-based WestJet Airlines Ltd. provides services throughout Canada and to some
international destinations. Excerpts from its 2013 financial statements are in Exhibits 2-19A to 2-
19C.
Exhibit 2-19A WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF
EARNINGS
Consolidated Statement of Earnings
For the years ended December 31
(Stated in thousands of Canadian dollars, except per share amounts)
Note
2013
2012
Revenue:
Guest
3,337,569
3,133,492
Other
324,628
293,917
3,662,197
3,427,409
Operating expenses:
Aircraft fuel
1,039,448
992,787
Airport operations
459,465
424,911
Flight operations and navigational charges
410,052
376,050
Sales and distribution
356,988
333,106
Marketing, general and administration
222,567
202,398
Depreciation and amortization
200,840
185,401
Inflight
176,907
162,633
Aircraft leasing
175,646
173,412
Maintenance
169,197
154,406
Employee profit share
51,577
46,585
3,262,687
3,051,689
Earnings from operations
399,510
375,720
Non-operating income (expense):
Finance income
15
17,848
18,391
Finance cost
15
(43,447)
(48,900)
Gain on foreign exchange
1,136
1,061
Gain (loss) on disposal of property and equipment
(2,962)
469
Loss on fuel derivatives
—
(6,512)
(27,425)
(35,491)
Earnings before income tax
372,085
340,229
Income tax expense (recovery):
Current
154,964
66,230
Deferred
(51,601)
31,607
11
103,363
97,837
Net earnings
268,722
242,392
Exhibit 2-19B WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF
FINANCIAL POSITION
Consolidated Statement of Financial Position
At December 31
(Stated in thousands of Canadian dollars)
Note
2013
2012
Assets
Current assets:
Cash and cash equivalents
5
1,256,005
1,408,199
Restricted cash
6
58,106
51,623
Accounts receivable
19
42,164
37,576
Prepaid expenses, deposits and other
19
133,263
101,802
Inventory
19
36,722
35,595
1,526,260
1,634,795
Non-current assets:
Property and equipment
7
2,487,734
1,985,599
Intangible assets
8
58,691
50,808
Other assets
19
70,778
75,413
Total assets
4,143,463
3,746,615
Liabilities and Shareholders\' Equity
Current liabilities:
Accounts payable and accrued liabilities
19
543,167
460,003
Advance ticket sales
19
551,022
480,947
Non-refundable guest credits
19
46,975
47,859
Current portion of maintenance provisions
9
76,105
34,135
Current portion of long-term debt
10
189,191
164,909
1,406,460
1,187,853
Non-current liabilities:
Maintenance provisions
9
142,411
145,656
Long-term debt
10
689,204
574,139
Other liabilities
19
8,834
9,914
.
Patient 2 Patient #2 Jan Johnson Next, Dr. Gupta sees Jan Johnson. .pdf
Patient 2 Patient #2: Jan Johnson Next, Dr. Gupta sees Jan Johnson. She is a 12-year-old female.
She is in the 7th grade and has been on the school swim team for the past four years. She is a
good student and is generally a happy child. Lately, Jan\'s mother has been worried about her
daughter. Jan has been very irritable the past few months and at first, her mother thought it was
related to the onset of puberty. In addition to mood changes, Mrs. Johnson has also noted a few
other changes in Jan she is often too tired to attend swimming practice and has complained of
feeling too weak to swim and of having headaches. Despite her decreased physical activity, Jan
has been eating almost non-stop but has lost about 5 lbs, she is also drinking more than usual and
has been vomiting frequently. Jan saw her primary care physician, and after running some tests,
has sent Jan to follow up with Dr. Gupta. Jan\'s blood work and patient information from his
chart are listed below.
Solution
1.Jan is easily fatigued, eating too much (polyphagia) drinking too much water (polydipsia),
sudden weight loss and nausea are the symptoms. All these symptoms relate to the metabolic
disorder called Diabetes.
2.jan is having high blood glucose level than what is considered to be normal. This indicates the
insulin receptors have become resistant to insulin which is type 2 diabetes or adequate amounts
of insulin is not being secreted by the pancreas which is type 1 diabetes. If untreated, diabetes
can be life threatening in the long term. It leads to diabetic retinopathy, kidney parenchymal
changes, coronary artet disease, diabetic foot etc.
3. I must collect information about her diet, frequency of urination, recent injuries, family history
of diabetes, any blurred vision,..
Non-math and physics question, but engineering orientated.Identify.pdf
Non-math and physics question, but engineering orientated.
Identify and discuss what you know are the significant milestones in materials used for
commercial aircraft since the end of the Second World War?
Include the manufacturing, maintenance and reliability aspects if relevant.
Limit your answer to three principal types of material.
Provide essay answers with a maximum of 500 words. Use current APA format for all
references. (minimum of 3 references) Thanks!
Solution
On December 17, 1903, Orville and Wilbur Wright capped four years of research and design
efforts with a 120-foot, 12-second flight at Kitty Hawk, North Carolina - the first powered flight
in a heavier-than-air machine. Prior to that, people had flown only in balloons and gliders. The
first person to fly as a passenger was Leon Delagrange, who rode with French pilot Henri
Farman from a meadow outside of Paris in 1908. Charles Furnas became the first American
airplane passenger when he flew with Orville Wright at Kitty Hawk later that year.
Aircraft Innovations
For the airlines to attract passengers away from the railroads, they needed both larger and faster
airplanes. They also needed safer airplanes. Accidents, such as the one in 1931 that killed Notre
Dame Football Coach Knute Rockne along with six others, kept people from flying
Aircraft manufacturers responded to the challenge. There were so many improvements to aircraft
in the 1930s that many believe it was the most innovative period in aviation history. Air-cooled
engines replaced water-cooled engines, reducing weight and making larger and faster planes
possible. Cockpit instruments also improved, with better altimeters, airspeed indicators, rate-of-
climb indicators, compasses, and the introduction of artificial horizon, which showed pilots the
attitude of the aircraft relative to the ground - important for flying in reduced visibility.
designers already knew that as an aircraft approaches the speed of sound (Mach 1), in the
transonic region, shock waves begin forming, causing a large increase in drag. Wings, already
thin, had to become thinner and finer. Fineness is a measure of how thin the wing is compared to
its front-to-back chord. A small, highly loaded wing has less drag and so some early types used
this type, including the Bell X-1 rocket plane and the Lockheed F-104 Starfighter. But these craft
had high takeoff speeds, the Starfighter causing significant pilot deaths during takeoff, and small
wings fell out of use. An approach pioneered by German designers during the war was to sweep
the wing at an angle, delaying the buildup of shock waves. But this made the wing structure
longer and more flexible, making the aircraft more likely to suffer from bending or aeroelasticity
and even causing a reversal in the action of the flight controls. Stall behaviour of the swept wing
was also poorly understood and could be extremely sharp. Other problems included divergent
oscillations which could build up lethal forces. In re.
Name three properties of the waveform that can be changed with the m.pdf
Name three properties of the waveform that can be changed with the multifunction
synthesizer/function generator.
Solution
Produce sine triangular sawtoothvwave form ..
*it can generate wide range of frequencies cover .1 Hz to 11 KHz
*frquency stability .1 % to analog generator and 550 ppm for digital generator
Output voltage can be 10 v peak to peak.
Inventory Valuation FIFO, LIFO, and Average The company reported the.pdf
Inventory Valuation: FIFO, LIFO, and Average The company reported the following inventory
data for the year: Practice 9-7 Units Cost per Unit 300 $1750 Beginning inventory Purchases:
900 1,200 400 18.00 18.25 March 23. Units remaining at year-end. . . . . . .. Compute (1) cost of
goods sold and (2) ending inventory assuming (a) FIFO inventory valuation, (b)LIFOin- ventory
valuation, and (c) average cost inventory valuation. The company uses a periodic inventory
system Practice 9-8 Inventory Valuation: Complications with a Perpetual System Refer to
Practice 9-7. Assume that the sales occurred as follows: Units Sold 100 600 1,300 July 15
November 1 Total Compute (1) cost of goods sold and (2) ending inventory assuming (a) FIFO
inventory valuation, (b) LIFO inven- tory valuation, and (c) average cost inventory valuation,
The company uses a perpetual inventory system
Solution
Answer
Units sold = 2400 – 400 = 2000
FIFO
Cost of Goods available for sale
Cost of Goods Sold
Endging Inventory
Units
Cost/unit
COG for sale
Units sold
Cost/unit
COGS
Units
Cost/unit
Ending inventory
Beginning Inventory
300
17.5
5250
300
17.5
5250
0
17.5
0
Purchases:
Mar-23
900
18
16200
900
18
16200
0
18
0
Sep-16
1200
18.25
21900
800
18.25
14600
400
18.25
7300
TOTAL
2400
43350
2000
36050
400
7300
LIFO
Cost of Goods available for sale
Cost of Goods Sold
Endging Inventory
Units
Cost/unit
COG for sale
Units sold
Cost/unit
COGS
Units
Cost/unit
Ending inventory
Beginning Inventory
300
17.5
5250
0
17.5
0
300
17.5
5250
Purchases:
Mar-23
900
18
16200
800
18
14400
100
18
1800
Sep-16
1200
18.25
21900
1200
18.25
21900
0
18.25
0
TOTAL
2400
43350
2000
36300
400
7050
---Average Cost Valuation
Cost of opening units + Purchases = $43350
Opening units + Purchased units = 2400
Average cost = 43350 / 2400 = $18.0625 per unit
Cost of Goods Sold = 2000 units sold x $18.0625 = $36125
Value of ending Inventory = 400 units x $18.0625 = $7225
FIFO
Opening + Purchases
Cost of Goods Sold
Closing Inventory
Opening units
300
17.50
5250
300
17.50
5250.00
Jan 16 Sale
100
17.50
1750
200
17.50
3500.00
Mar 23 Purchase
900
18.00
16200
200
17.50
3500.00
900
18.00
16200.00
July 15 Sale
200
17.50
3500
500
18.00
9000.00
400
18.00
7200
Sept 16 Purchase
1200
18.25
21900
500
18.00
9000.00
1200
18.25
21900.00
Nov 1 Sale
500
18.00
9000
400
18.25
7300.00
800
18.25
14600
TOTAL
2400
43350
2000
36050
400
7300.00
Units
Value
Cost of Goods Sold
2000
36050
Closing Inventory
400
7300
LIFO
Opening + Purchases
Cost of Goods Sold
Closing Inventory
Opening units
300
17.50
5250
300
17.50
5250.00
Jan 16 Sale
100
17.50
1750
200
17.50
3500.00
Mar 23 Purchase
900
18.00
16200
200
17.50
3500.00
900
18.00
16200.00
July 15 Sale
600
18.00
10800
200
17.50
3500.00
0
0.00
0
300
18.00
5400.00
Sept 16 Purchase
1200
18.25
21900
200
17.50
3500.00
300
18.00
5400.00
1200
18.25
21900.00
Nov 1 Sale
1200
18.25
21900
200
17.50
3500.00
100
18.00
1800
200
18.00
3600.00
TOTAL
2400
43350.00
2000
36250.00
400
7100
Units
Value
Cost of Goods Sold
2000
36250
.
In December 2009, a 45-year-old female presented to the emergency dep.pdf
In December 2009, a 45-year-old female presented to the emergency department (ED) 2 days
following abrupt onset of sore throat, nonproductive cough, chills, and mild fever. A chest
radiograph was performed, which was normal. She was diagnosed with bronchitis and asked to
follow up with her primary care physician, who subsequently started her on levofloxacin and
albuterol. Four days later she presented again to the ED with worsening cough, dispread, fever
(38.3 degree C; 101 degree F), and generalized lethargy. Additionally, she reported new
symptoms including a global headache, dizziness, myalgias, and arthralgias. She had no
abdominal pain, but reported nausea and anorexia. Her chest radiograph showed diffuse
reticulonodular opacities throughout the left lung, which were not present on her visit 4 days
previously. The patient was admitted for further evaluation. Questioning revealed the following:
she had a history of diabetes and hypertension. she smoked an average of a pack of cigarettes
daily, and she had received the seasonal influenza vaccine. Her husband was recently ill with
cough, but no order symptoms. On day 2 of hospitalization the patient\'s respiratory rate
increased from 22 to 46 and her oxygen saturation dropped while on oxygen administered by
nasal canola. She was transferred to an intensive care unit, where her respiratory status quickly
deteriorated, necessitating emergency intubation. A new chest radiograph showed bilateral
involvement, and she was begun on vancomycin, anreonam, and azithromycin. Blood cultures
drawn on her admission were negative, and an expectorated sputum sample taken at the same
time was not processed due to poor specimen quality. A PCR test performed on a
nasopharyngeal swab was positive for a viral agent, revealing the etiology of her infection (Fig.
10.1). What is the agent causing her infection? What are the key virulence factors of this agent?
How does this virus change over time? What made this virus unique in 2009?
Solution
Answer= 1] The agent causing her infection is influenza virus
Human influenza virus can evolve rapidly by 2 mechanisms
1] antigenic shift = genetic reassortment between a human & non human virus in non human
host
2] antigenic drift = accumulation of mutations that facilitate evasion of host immune response
New influenza viruses can be constantly emeges from environment
2] influenza constantly change by 2 different ways
antigenic drift = These are small changes in genes of influenza viruses that happen continuously
over time
as the virus replicates
Antigenic shift = is an abrupt major changes resulting in hemagglutinin & neuraminidase
proteins in influenza viruses that infect humans
Such a shift occured in 2009 when H1N1 virus with a new combination of genes emerged to
infect people & quickly spread causing a pandemic
when shift happens most people has no protection against H1N1.
In a longitudinal study of attitude toward statistics, a doctoral st.pdf
In a longitudinal study of attitude toward statistics, a doctoral student surveyed 10 doctoral
students at four times during their program: (a) before taking any statistics courses; (b) after
having taken one course; (c) after all required statistics courses; and (d) one semester after
completing their preliminary examination. In each instance, alternate forms of the Attitude
Towards Statistics instrument were used. The data were:
1. Analyze the data using an alpha of .05 in SPSS (Free trial of SPSS here: http://www-
01.ibm.com/software/analytics/spss/products/statistics/downloads.html)
2. Writing the result in APA format.
Solution
SPSS out put
GLM v1 v2 v3 v4
/WSFACTOR=factor1 4 Polynomial
/METHOD=SSTYPE(3)
/EMMEANS=TABLES(factor1) COMPARE ADJ(BONFERRONI)
/PRINT=DESCRIPTIVE
/CRITERIA=ALPHA(.05)
/WSDESIGN=factor1.
Descriptive Statistics
Mean
Std. Deviation
N
v1
40.50
11.008
10
v2
42.40
11.806
10
v3
43.00
11.126
10
v4
36.80
10.390
10
Multivariate Testsa
Effect
Value
F
Hypothesis df
Error df
Sig.
factor1
Pillai\'s Trace
.768
7.725b
3.000
7.000
.013
Wilks\' Lambda
.232
7.725b
3.000
7.000
.013
Hotelling\'s Trace
3.311
7.725b
3.000
7.000
.013
Roy\'s Largest Root
3.311
7.725b
3.000
7.000
.013
a. Design: Intercept
Within Subjects Design: factor1
b. Exact statistic
Mauchly\'s Test of Sphericitya
Measure: MEASURE_1
Within Subjects Effect
Mauchly\'s W
Approx. Chi-Square
df
Sig.
Epsilonb
Greenhouse-Geisser
Huynh-Feldt
Lower-bound
factor1
.509
5.209
5
.394
.688
.896
.333
Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed
dependent variables is proportional to an identity matrix.
a. Design: Intercept
Within Subjects Design: factor1
b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected
tests are displayed in the Tests of Within-Subjects Effects table.
Tests of Within-Subjects Effects
Measure: MEASURE_1
Source
Type III Sum of Squares
df
Mean Square
F
Sig.
factor1
Sphericity Assumed
234.275
3
78.092
15.916
.000
Greenhouse-Geisser
234.275
2.064
113.520
15.916
.000
Huynh-Feldt
234.275
2.687
87.190
15.916
.000
Lower-bound
234.275
1.000
234.275
15.916
.003
Error(factor1)
Sphericity Assumed
132.475
27
4.906
Greenhouse-Geisser
132.475
18.574
7.132
Huynh-Feldt
132.475
24.182
5.478
Lower-bound
132.475
9.000
14.719
Tests of Within-Subjects Contrasts
Measure: MEASURE_1
Source
factor1
Type III Sum of Squares
df
Mean Square
F
Sig.
factor1
Linear
55.125
1
55.125
20.066
.002
Quadratic
164.025
1
164.025
18.401
.002
Cubic
15.125
1
15.125
4.946
.053
Error(factor1)
Linear
24.725
9
2.747
Quadratic
80.225
9
8.914
Cubic
27.525
9
3.058
Tests of Between-Subjects Effects
Measure: MEASURE_1
Transformed Variable: Average
Source
Type III Sum of Squares
df
Mean Square
F
Sig.
Intercept
66178.225
1
66178.225
138.576
.000
Error
4298.025
9
477.558
Estimated Marginal Means
factor1
Estimates
Measure: MEASURE_1
factor1
Mean
Std. Error
95% Confidence Interval
Lower Bound
Upper Bound
1
40.500
3.481
32.626
48..
Introduction to AI for Nonprofits with Tapp Network
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
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According to the IRS, 1.1 of tax returns will be audited in 2011. A.pdf
According to the IRS, 1.1% of tax returns will be audited in 2011. A random sample of 40 tax
returns was selected. Use the Poisson distribution to approximate the binomial distribution to
determine the probability that more than two of these tax returns will be audited.0.1215
Solution
for poisson distribution, mean = np = 40(0.011) = 0.44
by using excel function =1-POISSON(1,0.44,1), P(x>=2) = 0.0726.
Write a SELECT statement that returns a single value that represents.pdf
Write a SELECT statement that returns a single value that represents the sum of the largest
unpaid invoices submitted by each vendor. Use a common table expression(cte) that returns
MAX(InvoiceTotal) grouped by VendorID, filtering for invoices with a balanced due.
Solution
WITH INVOICE_CTE (VendorID,maxInvoiceTotal)
AS
(
Select VendorID,MAX(InvoiceTotal) as maxInvoiceTotal
from Invoice
where due>0
)
Select MAX(maxInvoiceTotal) from
INVOICE_CTE.
which represents the ground state for the N- ion i. w.ia ere sm.pdf
which represents the ground state for the N- ion
i. w.ia ere sms he Morral su rrie ...dae ... lilili
Solution
The ground state for the N- ion
option e..
The ground state of an atom or ion is the state in which all its electrons are found in the possible
lowest energy orbitals .
your rest of the image of questions are not clear ..
What is the ecological and environmental importance of salinityoxyg.pdf
What is the ecological and environmental importance of salinity/oxygen consumption of grass
shrimp?
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Grass shrimp acts as detritivores,grass shrimp aid in the mechanical break down of organic
matter such as fibrous plant materials into tiny particles that are suspended in the water column,
and thus provides rich food source for smaller organisms. Predation : In estuaries, numerous fish
species and other aquatic carnivores , eat large quantities of grass shrimp. Grass shrimp are hosts
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WEP has vulnerabilities. Which of the following is not a reason why .pdf
WEP has vulnerabilities. Which of the following is not a reason why it is vulnerable?
A. Shared WEP keys among all clients
B. An RC4 engine not properly initialized
C. 20-bit initialization vector
D. 40-bit WEP keys
Solution
C. One issue with WEP is the initialization vector (IV), it was 24 bits, not 20. Answers A, B, and
D detail some of the vulnerabilities of WEP. For example, WEP uses a single shared key among
all clients, which means that you are authenticating groups, not devices or single users. Also,
RC4 is the correct encryption type and can be implemented in 40- or 104-bit configuration, but
WEP does not properly initialize it. This means that the key values roll over and are predictable.
Finally, a 24-bit IV vector is too short, and a 40-bit key is weak..
Two astronauts are 2.40 m apart in their spaceship. One speaks to the.pdf
Two astronauts are 2.40 m apart in their spaceship. One speaks to the other. The conversation is
transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at 343
m/s through the air between the astronauts equals the time it takes for the electromagnetic waves
to travel to the earth. How far away from the earth is the spaceship Number Units
Solution
time taken by sound to travel 2.40 m,
t = 2.40 / 343 s
spaceship is at the distance at light can travel in time t.
distance = speed x time
= (3 x 10^8 ) (2.40 / 343)
= 2.099 x 10^6 m.
The ratio of the probability of disease in an exposed group to the p.pdf
The ratio of the probability of disease in an exposed group to the probability of disease in an
unexposed group is called the: 1. Relative risk 2. Odds ratio 3. Risk difference 4. Attributable
risk
Solution
The ratio of the probability of disease in an exposed group to the probability of disease in an
unexposed group is called relative risk.
RR=P(diseased/exposed)/P(diseased/unexposed).
The output should now look like this Row 1 sum 30 Row 2 sum.pdf
The output should now look like this:
Row 1 sum: 30
Row 2 sum: 48
Row 3 sum: 55
Count: 12
Grand total: 133
Mean: 11.08
Even numbers: 7
Odd numbers: 5
How to add even and odd numbers counts to existing code
public class Problem {
public static void main(String [] args)
{
int iVal;
int iRow;
int iRowSum;
int iToken;
int totalSum = 0;
int totalEle = 0;
iRowSum = 0;
iRow = 1;
try
{
Scanner ifsInput = new Scanner(new File(\"c:\\home\\student\\lastname/9_1_Input.txt\"));
while(ifsInput.hasNextLine())
{
while(ifsInput.hasNextInt())
{
for(iToken = 0; iToken <= 3; iToken++)
{
iVal = ifsInput.nextInt();
iRowSum = iRowSum + iVal;
// increasing count of total element
totalEle++;
}
// adding into totalSum
totalSum = totalSum + iRowSum;
System.out.println(\"Row \" + iRow + \" sum: \" + iRowSum);
iRowSum = 0;
iRow++;
}
}
// added new code to print grand total and mean value
System.out.println(\"Count: \"+totalEle);
System.out.println(\"Grand total: \"+totalSum);
System.out.println(\"Mean: \"+String.format(\".2f\", (double)/totalSum/totalEle);
}
catch (FileNotFoundException sMsg)
{
System.out.println(\"The file cannot be found or opened.\");
}
}
}
Solution
Input999.txt( Save this file under D Drive Then the path of the file pointing to it is
D:\\\\Input999.txt)
11 22 33 44
22 33 44 55
33 44 55 66
___________________________
Problem.java
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Problem {
public static void main(String [] args)
{
//Declaring variables
int iVal;
int iRow;
int iRowSum;
int iToken;
double mean=0.0;
int evenCount=0,oddCount=0;
int totalSum = 0;
int totalEle = 0;
iRowSum = 0;
iRow = 1;
try
{
Scanner ifsInput = new Scanner(new File(\"D:\\\\Input999.txt\"));
while(ifsInput.hasNextLine())
{
while(ifsInput.hasNextInt())
{
for(iToken = 0; iToken <= 3; iToken++)
{
iVal = ifsInput.nextInt();
//Calculating the even count
if(iVal%2==0)
evenCount++;
//Calculating the odd count
else
oddCount++;
//Calculating the sum of elements in each row
iRowSum = iRowSum + iVal;
// increasing count of total element
totalEle++;
}
// adding into totalSum
totalSum = totalSum + iRowSum;
//Displaying the sum of elements in each row
System.out.println(\"Row \" + iRow + \" sum: \" + iRowSum);
iRowSum = 0;
iRow++;
}
}
//Displaying the number of elements
System.out.println(\"Count: \"+totalEle);
//Displaying the sum of all elements
System.out.println(\"Grand total: \"+totalSum);
//Calculating the mean
mean=(double)totalSum/totalEle;
//Displaying the mean
System.out.printf(\"Mean: %.2f\ \",mean);
//Displaying the number of even nos
System.out.println(\"Even Numbers :\"+evenCount);
//Displaying the number of odd nos
System.out.println(\"Odd Numbers :\"+oddCount);
}
catch (FileNotFoundException sMsg)
{
System.out.println(\"The file cannot be found or opened.\");
}
}
}
_________________________________
Output:
Row 1 sum: 110
Row 2 sum: 154
Row 3 sum: 198
Count: 12
Grand total: 462
Mean: 38.50
Even Numbers :6
Odd Numbers :6
______.
The end problem in eukaryotic DNA replicationa. is solved by t.pdf
The \"end problem\" in eukaryotic DNA replication
a. is solved by telomerase
b. refers to loss of chromosome ends during replication
c. is solved by an RNA dependent DNA polymerase
d. A and B
e. all of the above
Solution
Answer : The correct answer is E ( All of the above)
Explanation :
Telomerases are the ribonucleoprotein complexes containing a small RNA that serves as a
template for addition of new six nucleotide repeats. Or Telomerase, the enzyme that synthesizes
the telomere ends of linear chromosomes, is a specialized reverse transcriptase that contains an
internal RNA template. Telomeres undergo cycles of shortening of the lagging strands due to the
inability to complete synthesis.
Most normal cells lose telomerase activity with increaseing age of the organism and repeated
generations of the cell division. In other words, after certain number of divisions normal cells
reaches the Hayflicks limit beyond which cells will no longer devide instead they die.
In germ-line cells, which contain telomerase activity, telomere lengths are maintained; for
example, telomerase is highly expressed only in cells that need to divide regularly especially in
male sperm cells. Whereas somatic cells lack telomerase activity due to which they die after
certain number of generations.
In cancer cells/tumor cells, end replication problems observed in normal cells will be taken care
by the telomerase enzyme which makes the cells to bypass the Hayflick limit and become
immortal, thus avoiding cell death Many cancer cells are considered \'immortal\' because
telomerase activity supports them to live longer than somatic cells. 90% of the cancer cells retain
the ability to express telomerase and acquires the ability to proloferate indefinitely..
Suppose that one obtained the following DNA sequence data for the fou.pdf
Suppose that one obtained the following DNA sequence data for the four fly families and a non-
fly insect (the outgroup) from a gene that is not related to morphology. We will call the DNA
sequence positions 1 through 10, for reference (in fly family A, position 1 is a T, while position
10 is an A). Which of the characters (nucleotide positions) are variable among the fly families
(not counting the outgroup? Which of the characters (nucleotide positions) are phylogenetically
informative for cladistics? (Phylogenetically informative traits are those for which there is
variation in the group of interest, and for which there is evidence of synapomorphies.)
Solution
Purines are 2 ringed structure= A,G whereas, pyramidines aresingle ringed structure = C,T
Generally A = T (A double bonds with T) and G=C (G triple H bonds with C).
Mutation causes a change in theDNA sequence. Two types
1. Transversions: Purine instead of a pyramidine or vice versa (Less frequent so more important
for cladistics) A/G to C/T or vice versa. 2 ringed structure to 1 ringed structure
See this is seen in all fly families at 7th(ABCD:AGGC) , 8th (ABCD:CTAC) and 9th
(ABCD:GCTG)positions
2. Transitions: Purine instead of a puirne and pyramidine instead of a pyramidine (Easy and
frequent)
The Nucleotide positions are Bolded in the table are variable among the fly families. Among all
Fly family C is most variable. (Take A, B, C, D and leave Out group).
Phylogenetically informative traits are those for which there is variation in the group of interest
and for which there is evidence of synapomorphies. All this is to know the members of a group
which share a common history and to know their closely related species. Synapomorphy means
any characteristic which is present in an ancestral species and that is shared exclusively (either in
more or less modified form) by its evolutionary descendants.
Mostly there is a much variation in the 7th, 8th and 9th position of the nucleotide in all fly
families sequence so this will be more phylogenetically informative for cladistics and the 5th
sequence is the most stable one. From the sequence fly family C is much more related to the out
group sequence than all familiesFLY FAMILYDNA Sequence alignedpositions
order12345678910ATGCGTTACGABTGCATTGTCACTCCGTAGATTDGCAGTTCCGAOU
T GROUPGCAGTTGATT.
Set up a JavaFX GUI-based program that shows a 10 times 10 grid of la.pdf
Set up a JavaFX GUI-based program that shows a 10 times 10 grid of labels that forms a
multiplication table, with the labels displaying the multiplication problems, rather than the
answers. Provide a text input field and a button with an event handler that reads an integer value
from the text input and changes the CSS styling in some obvious way for all problems in the
table with the given answer. When a new answer is entered and the button is clicked, change all
the labels back to the original style, then change the labels showing problems for the new answer
to the new style. Use CSS, not setters, for all the styling. The main point of this lab is to learn
JavaFX, so the grading will include a heavy emphasis on the neatness and general appearance of
your GUI.
Solution
Code:
//multiTable.java
// import required packages
import java.util.List;
import javafx.application.Application;
import javafx.event.Event;
import javafx.event.EventHandler;
import javafx.geometry.Insets;
import javafx.geometry.Pos;
import javafx.scene.Node;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.control.TextField;
import javafx.scene.input.MouseEvent;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.HBox;
import javafx.stage.Stage;
// class for revrse multiplication table
public class multiTable extends Application
{
// over ride nethod for setup the grid
@Override
public void start(Stage primaryStage)
{
BorderPane pane1 = new BorderPane();
pane1.setTop(getHbox1());
HBox h_box = new HBox(15);
h_box.setPadding(new Insets(15, 15, 15, 15));
h_box.setAlignment(Pos.TOP_CENTER);
h_box.getStyleClass().add(\"hbox2\");
Label labl = new Label(\"Enter Answer: \");
h_box.getChildren().add(labl);
TextField textfield1 = new TextField();
h_box.getChildren().add(textfield1);
GridPane grid_pane1 = setUpGrid();
gridmaker griddp = new gridmaker(grid_pane1);
Button Answer = new Button(\"Find problems\");
Answer.addEventHandler(MouseEvent.MOUSE_CLICKED, new EventHandler()
{
@Override
public void handle(Event arg0)
{
List fact = factors(textfield1);
for (int[] RCx1 : fact) {
Node node = griddp.get_chldren()[RCx1[0]][RCx1[1]];
node.setStyle(\"-fx-background-color: green\");
}
}
});
h_box.getChildren().add(Answer);
pane1.setCenter(h_box);
pane1.setBottom(grid_pane1);
Scene scene = new Scene(pane1, 550, 650);
scene.getStylesheets().add(getClass().getResource(\"application.css\").toExternalForm());
primaryStage.setTitle(\"lab 6\");
primaryStage.setScene(scene);
primaryStage.show();
}
// method for grid layout
private HBox getHbox1()
{
HBox hbox2 = new HBox(15);
hbox2.setPadding(new Insets(15, 15, 15, 15));
hbox2.setAlignment(Pos.TOP_CENTER);
hbox2.getStyleClass().add(\"hbox1\");
Label labl = new Label(\"Multiplication Table\");
hbox2.getChildren().add(labl);
return hbox2;
}
// method to setup a grid
public GridPane setUpGrid() {
GridPane pane1 = new GridPane();
Label[][].
Sample of DNA isolated from bacterium X contains 17.5 of adenine.pdf
Sample of DNA isolated from bacterium X contains 17.5% of adenine bases. What percentages
of guanine, thymine, and cytosine bases would you expect to find in the same sample of DNA?
Match the following types of transport across a biological membrane (designated I, II, or III)
with the statements designated A-D: Simple diffusion (SD) Facilitated diffusion (FD) Active
transport (AT) Write down your answers in the blank space next to each statement below. Note,
that more than one answer may apply each statement: Process is energy dependent: ___ Process
needs transmembrane transporter/channel: ___ Process occurs down the concentration gradient:
___ Processes can establish a concentration gradient: ___
Solution
Answer 2. Given that adenine bases in the isolated DNA of the bacterium is 17.5%. Since
Adenine is known to pair with thymine. So, the percentage of thymine will also be 17.5%.
Adenine and thymine will make 35 % of the total nitrogen bases of the DNA. Rest of the DNA
will be made from Guanine and Cytosine. Guanine and Cytosine are known to pair with each
other. So, their content in the DNA will also be same. 100-35% will give us 65 %. So Guanine
will be 32.5% and Cytosine will be 32.5%
Answer 3.
Process occur down the concentration gradient - Simple Diffusion(SD)
Process need transmembrane transporter/Channel- Facilitated Diffusion (FD)
Process is energy dependent- Active transport (AT)
Process can establish a concentration gradient- Facilitated Diffusion (FD) and Active transport
(AT).
Practice using layouts Anderson, Franceschi import javax..pdf
/* Practice using layouts
Anderson, Franceschi
*/
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
public class NestedLayoutPractice extends JFrame
{
private Container contents;
private Game game;
private BorderLayout bl;
private JLabel bottom;
// ***** Task 1: declare a JPanel named top
// also declare three JButton instance variables
// that will be added to the JPanel top
// these buttons will determine the grid size of the game:
// 3-by-3, 4-by-4, or 5-by-5
// Part 1 student code starts here:
private final JPanel top;
private JButton [] topButtones;
String [] topGridSize =
{\"3-by-3\",
\"4-by-4\",
\"5-by-5\",
};
private JButton reset;
// Part 1 student code ends here.
public NestedLayoutPractice()
{
super(\"Practicing layout managers\");
contents = getContentPane();
// ***** Task 2:
// instantiate the BorderLayout manager bl
// Part 2 student code starts here:
// set the layout manager of the content pane contents to bl:
// bl = new BorderLayout (); (not suer if this is right)
//set.BorderLayout(bl); (not suer if this is right)
game = new Game(3); // instantiating the GamePanel object
// add panel (game) to the center of the content pane
// Part 2 student code ends here.
bottom = new JLabel(\"Have fun playing this Tile Puzzle game\",
SwingConstants.CENTER);
// ***** Task 3:
// instantiate the JPanel component named top
// Part 3 student code starts here:
//top = new JPanle ();(not suer if this is right)
// set the layout of top to a 1-by-3 grid(not suer if this is right)
// panel.setLayout (new GridLayout (3, 1));(not suer if this is right)
// instantiate the JButtons that determine the grid size
// topButtones [0] = new JButton (\"3-by-3 \"); (not suer if this is right)
// add the buttons to JPanel top
//top.add (topButtones[0]);(not suer if this is right)
//top.add (topButtones[1]);
//top.add (topButtones[2]);
// add JPanel top to the content pane as its north component
c.add(top, \"north\");
// Part 3 student code ends here.
// ***** Task 5:
// Note: search for and complete Task 4 before performing this task
// Part 5 student code starts here:
// declare and instantiate an ActionListener
//topButtones[0].addActionListnere(this);(not suer if this is right)
//topButtones[1].addActionListnere(this);
//topButtones[2].addActionListnere(this);
// register the listener on the 3 buttons
// that you declared in Task 1
// Part 5 student code ends here.
contents.add(bottom, BorderLayout.SOUTH);
setSize(325, 325);
setVisible(true);
}
// ***** Task 4:
// create a private inner class that implements ActionListener
// your method should identify which of the 3 buttons
// was the source of the event
// depending on which button was pressed,
// call the setUpGame method of the Game class
// with arguments 3, 4, or 5
// the API of that method is:
// public void setUpGame(int nSides)
// At the end of the method call validate()
// Part 4 student code starts here:
public void actionPreformed(ActionEvent ae)
{
if (ae.getActionCommand(). equ.
Please, I need a correct answer and clear explanation. open image an.pdf
Please, I need a correct answer and clear explanation. open image and zoom for a clear view.
Thanks,
Using the West Jet Financial statements, answer the following questions R1 2-4 p.106-109)
b. Why do you think West Jet does not report a cost of goods sold
c. Explain what “Advanced ticket sales” in the current liabilities section represents
e. How much did West Jet use in 2013 to acquire new aircraft? Which statement did you find this
information on?
RI2-4
(Determination of items from a Canadian company\'s financial statements)
Calgary-based WestJet Airlines Ltd. provides services throughout Canada and to some
international destinations. Excerpts from its 2013 financial statements are in Exhibits 2-19A to 2-
19C.
Exhibit 2-19A WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF
EARNINGS
Consolidated Statement of Earnings
For the years ended December 31
(Stated in thousands of Canadian dollars, except per share amounts)
Note
2013
2012
Revenue:
Guest
3,337,569
3,133,492
Other
324,628
293,917
3,662,197
3,427,409
Operating expenses:
Aircraft fuel
1,039,448
992,787
Airport operations
459,465
424,911
Flight operations and navigational charges
410,052
376,050
Sales and distribution
356,988
333,106
Marketing, general and administration
222,567
202,398
Depreciation and amortization
200,840
185,401
Inflight
176,907
162,633
Aircraft leasing
175,646
173,412
Maintenance
169,197
154,406
Employee profit share
51,577
46,585
3,262,687
3,051,689
Earnings from operations
399,510
375,720
Non-operating income (expense):
Finance income
15
17,848
18,391
Finance cost
15
(43,447)
(48,900)
Gain on foreign exchange
1,136
1,061
Gain (loss) on disposal of property and equipment
(2,962)
469
Loss on fuel derivatives
—
(6,512)
(27,425)
(35,491)
Earnings before income tax
372,085
340,229
Income tax expense (recovery):
Current
154,964
66,230
Deferred
(51,601)
31,607
11
103,363
97,837
Net earnings
268,722
242,392
Exhibit 2-19B WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF
FINANCIAL POSITION
Consolidated Statement of Financial Position
At December 31
(Stated in thousands of Canadian dollars)
Note
2013
2012
Assets
Current assets:
Cash and cash equivalents
5
1,256,005
1,408,199
Restricted cash
6
58,106
51,623
Accounts receivable
19
42,164
37,576
Prepaid expenses, deposits and other
19
133,263
101,802
Inventory
19
36,722
35,595
1,526,260
1,634,795
Non-current assets:
Property and equipment
7
2,487,734
1,985,599
Intangible assets
8
58,691
50,808
Other assets
19
70,778
75,413
Total assets
4,143,463
3,746,615
Liabilities and Shareholders\' Equity
Current liabilities:
Accounts payable and accrued liabilities
19
543,167
460,003
Advance ticket sales
19
551,022
480,947
Non-refundable guest credits
19
46,975
47,859
Current portion of maintenance provisions
9
76,105
34,135
Current portion of long-term debt
10
189,191
164,909
1,406,460
1,187,853
Non-current liabilities:
Maintenance provisions
9
142,411
145,656
Long-term debt
10
689,204
574,139
Other liabilities
19
8,834
9,914
.
Patient 2 Patient #2 Jan Johnson Next, Dr. Gupta sees Jan Johnson. .pdf
Patient 2 Patient #2: Jan Johnson Next, Dr. Gupta sees Jan Johnson. She is a 12-year-old female.
She is in the 7th grade and has been on the school swim team for the past four years. She is a
good student and is generally a happy child. Lately, Jan\'s mother has been worried about her
daughter. Jan has been very irritable the past few months and at first, her mother thought it was
related to the onset of puberty. In addition to mood changes, Mrs. Johnson has also noted a few
other changes in Jan she is often too tired to attend swimming practice and has complained of
feeling too weak to swim and of having headaches. Despite her decreased physical activity, Jan
has been eating almost non-stop but has lost about 5 lbs, she is also drinking more than usual and
has been vomiting frequently. Jan saw her primary care physician, and after running some tests,
has sent Jan to follow up with Dr. Gupta. Jan\'s blood work and patient information from his
chart are listed below.
Solution
1.Jan is easily fatigued, eating too much (polyphagia) drinking too much water (polydipsia),
sudden weight loss and nausea are the symptoms. All these symptoms relate to the metabolic
disorder called Diabetes.
2.jan is having high blood glucose level than what is considered to be normal. This indicates the
insulin receptors have become resistant to insulin which is type 2 diabetes or adequate amounts
of insulin is not being secreted by the pancreas which is type 1 diabetes. If untreated, diabetes
can be life threatening in the long term. It leads to diabetic retinopathy, kidney parenchymal
changes, coronary artet disease, diabetic foot etc.
3. I must collect information about her diet, frequency of urination, recent injuries, family history
of diabetes, any blurred vision,..
Non-math and physics question, but engineering orientated.Identify.pdf
Non-math and physics question, but engineering orientated.
Identify and discuss what you know are the significant milestones in materials used for
commercial aircraft since the end of the Second World War?
Include the manufacturing, maintenance and reliability aspects if relevant.
Limit your answer to three principal types of material.
Provide essay answers with a maximum of 500 words. Use current APA format for all
references. (minimum of 3 references) Thanks!
Solution
On December 17, 1903, Orville and Wilbur Wright capped four years of research and design
efforts with a 120-foot, 12-second flight at Kitty Hawk, North Carolina - the first powered flight
in a heavier-than-air machine. Prior to that, people had flown only in balloons and gliders. The
first person to fly as a passenger was Leon Delagrange, who rode with French pilot Henri
Farman from a meadow outside of Paris in 1908. Charles Furnas became the first American
airplane passenger when he flew with Orville Wright at Kitty Hawk later that year.
Aircraft Innovations
For the airlines to attract passengers away from the railroads, they needed both larger and faster
airplanes. They also needed safer airplanes. Accidents, such as the one in 1931 that killed Notre
Dame Football Coach Knute Rockne along with six others, kept people from flying
Aircraft manufacturers responded to the challenge. There were so many improvements to aircraft
in the 1930s that many believe it was the most innovative period in aviation history. Air-cooled
engines replaced water-cooled engines, reducing weight and making larger and faster planes
possible. Cockpit instruments also improved, with better altimeters, airspeed indicators, rate-of-
climb indicators, compasses, and the introduction of artificial horizon, which showed pilots the
attitude of the aircraft relative to the ground - important for flying in reduced visibility.
designers already knew that as an aircraft approaches the speed of sound (Mach 1), in the
transonic region, shock waves begin forming, causing a large increase in drag. Wings, already
thin, had to become thinner and finer. Fineness is a measure of how thin the wing is compared to
its front-to-back chord. A small, highly loaded wing has less drag and so some early types used
this type, including the Bell X-1 rocket plane and the Lockheed F-104 Starfighter. But these craft
had high takeoff speeds, the Starfighter causing significant pilot deaths during takeoff, and small
wings fell out of use. An approach pioneered by German designers during the war was to sweep
the wing at an angle, delaying the buildup of shock waves. But this made the wing structure
longer and more flexible, making the aircraft more likely to suffer from bending or aeroelasticity
and even causing a reversal in the action of the flight controls. Stall behaviour of the swept wing
was also poorly understood and could be extremely sharp. Other problems included divergent
oscillations which could build up lethal forces. In re.
Name three properties of the waveform that can be changed with the m.pdf
Name three properties of the waveform that can be changed with the multifunction
synthesizer/function generator.
Solution
Produce sine triangular sawtoothvwave form ..
*it can generate wide range of frequencies cover .1 Hz to 11 KHz
*frquency stability .1 % to analog generator and 550 ppm for digital generator
Output voltage can be 10 v peak to peak.
Inventory Valuation FIFO, LIFO, and Average The company reported the.pdf
Inventory Valuation: FIFO, LIFO, and Average The company reported the following inventory
data for the year: Practice 9-7 Units Cost per Unit 300 $1750 Beginning inventory Purchases:
900 1,200 400 18.00 18.25 March 23. Units remaining at year-end. . . . . . .. Compute (1) cost of
goods sold and (2) ending inventory assuming (a) FIFO inventory valuation, (b)LIFOin- ventory
valuation, and (c) average cost inventory valuation. The company uses a periodic inventory
system Practice 9-8 Inventory Valuation: Complications with a Perpetual System Refer to
Practice 9-7. Assume that the sales occurred as follows: Units Sold 100 600 1,300 July 15
November 1 Total Compute (1) cost of goods sold and (2) ending inventory assuming (a) FIFO
inventory valuation, (b) LIFO inven- tory valuation, and (c) average cost inventory valuation,
The company uses a perpetual inventory system
Solution
Answer
Units sold = 2400 – 400 = 2000
FIFO
Cost of Goods available for sale
Cost of Goods Sold
Endging Inventory
Units
Cost/unit
COG for sale
Units sold
Cost/unit
COGS
Units
Cost/unit
Ending inventory
Beginning Inventory
300
17.5
5250
300
17.5
5250
0
17.5
0
Purchases:
Mar-23
900
18
16200
900
18
16200
0
18
0
Sep-16
1200
18.25
21900
800
18.25
14600
400
18.25
7300
TOTAL
2400
43350
2000
36050
400
7300
LIFO
Cost of Goods available for sale
Cost of Goods Sold
Endging Inventory
Units
Cost/unit
COG for sale
Units sold
Cost/unit
COGS
Units
Cost/unit
Ending inventory
Beginning Inventory
300
17.5
5250
0
17.5
0
300
17.5
5250
Purchases:
Mar-23
900
18
16200
800
18
14400
100
18
1800
Sep-16
1200
18.25
21900
1200
18.25
21900
0
18.25
0
TOTAL
2400
43350
2000
36300
400
7050
---Average Cost Valuation
Cost of opening units + Purchases = $43350
Opening units + Purchased units = 2400
Average cost = 43350 / 2400 = $18.0625 per unit
Cost of Goods Sold = 2000 units sold x $18.0625 = $36125
Value of ending Inventory = 400 units x $18.0625 = $7225
FIFO
Opening + Purchases
Cost of Goods Sold
Closing Inventory
Opening units
300
17.50
5250
300
17.50
5250.00
Jan 16 Sale
100
17.50
1750
200
17.50
3500.00
Mar 23 Purchase
900
18.00
16200
200
17.50
3500.00
900
18.00
16200.00
July 15 Sale
200
17.50
3500
500
18.00
9000.00
400
18.00
7200
Sept 16 Purchase
1200
18.25
21900
500
18.00
9000.00
1200
18.25
21900.00
Nov 1 Sale
500
18.00
9000
400
18.25
7300.00
800
18.25
14600
TOTAL
2400
43350
2000
36050
400
7300.00
Units
Value
Cost of Goods Sold
2000
36050
Closing Inventory
400
7300
LIFO
Opening + Purchases
Cost of Goods Sold
Closing Inventory
Opening units
300
17.50
5250
300
17.50
5250.00
Jan 16 Sale
100
17.50
1750
200
17.50
3500.00
Mar 23 Purchase
900
18.00
16200
200
17.50
3500.00
900
18.00
16200.00
July 15 Sale
600
18.00
10800
200
17.50
3500.00
0
0.00
0
300
18.00
5400.00
Sept 16 Purchase
1200
18.25
21900
200
17.50
3500.00
300
18.00
5400.00
1200
18.25
21900.00
Nov 1 Sale
1200
18.25
21900
200
17.50
3500.00
100
18.00
1800
200
18.00
3600.00
TOTAL
2400
43350.00
2000
36250.00
400
7100
Units
Value
Cost of Goods Sold
2000
36250
.
In December 2009, a 45-year-old female presented to the emergency dep.pdf
In December 2009, a 45-year-old female presented to the emergency department (ED) 2 days
following abrupt onset of sore throat, nonproductive cough, chills, and mild fever. A chest
radiograph was performed, which was normal. She was diagnosed with bronchitis and asked to
follow up with her primary care physician, who subsequently started her on levofloxacin and
albuterol. Four days later she presented again to the ED with worsening cough, dispread, fever
(38.3 degree C; 101 degree F), and generalized lethargy. Additionally, she reported new
symptoms including a global headache, dizziness, myalgias, and arthralgias. She had no
abdominal pain, but reported nausea and anorexia. Her chest radiograph showed diffuse
reticulonodular opacities throughout the left lung, which were not present on her visit 4 days
previously. The patient was admitted for further evaluation. Questioning revealed the following:
she had a history of diabetes and hypertension. she smoked an average of a pack of cigarettes
daily, and she had received the seasonal influenza vaccine. Her husband was recently ill with
cough, but no order symptoms. On day 2 of hospitalization the patient\'s respiratory rate
increased from 22 to 46 and her oxygen saturation dropped while on oxygen administered by
nasal canola. She was transferred to an intensive care unit, where her respiratory status quickly
deteriorated, necessitating emergency intubation. A new chest radiograph showed bilateral
involvement, and she was begun on vancomycin, anreonam, and azithromycin. Blood cultures
drawn on her admission were negative, and an expectorated sputum sample taken at the same
time was not processed due to poor specimen quality. A PCR test performed on a
nasopharyngeal swab was positive for a viral agent, revealing the etiology of her infection (Fig.
10.1). What is the agent causing her infection? What are the key virulence factors of this agent?
How does this virus change over time? What made this virus unique in 2009?
Solution
Answer= 1] The agent causing her infection is influenza virus
Human influenza virus can evolve rapidly by 2 mechanisms
1] antigenic shift = genetic reassortment between a human & non human virus in non human
host
2] antigenic drift = accumulation of mutations that facilitate evasion of host immune response
New influenza viruses can be constantly emeges from environment
2] influenza constantly change by 2 different ways
antigenic drift = These are small changes in genes of influenza viruses that happen continuously
over time
as the virus replicates
Antigenic shift = is an abrupt major changes resulting in hemagglutinin & neuraminidase
proteins in influenza viruses that infect humans
Such a shift occured in 2009 when H1N1 virus with a new combination of genes emerged to
infect people & quickly spread causing a pandemic
when shift happens most people has no protection against H1N1.
In a longitudinal study of attitude toward statistics, a doctoral st.pdf
In a longitudinal study of attitude toward statistics, a doctoral student surveyed 10 doctoral
students at four times during their program: (a) before taking any statistics courses; (b) after
having taken one course; (c) after all required statistics courses; and (d) one semester after
completing their preliminary examination. In each instance, alternate forms of the Attitude
Towards Statistics instrument were used. The data were:
1. Analyze the data using an alpha of .05 in SPSS (Free trial of SPSS here: http://www-
01.ibm.com/software/analytics/spss/products/statistics/downloads.html)
2. Writing the result in APA format.
Solution
SPSS out put
GLM v1 v2 v3 v4
/WSFACTOR=factor1 4 Polynomial
/METHOD=SSTYPE(3)
/EMMEANS=TABLES(factor1) COMPARE ADJ(BONFERRONI)
/PRINT=DESCRIPTIVE
/CRITERIA=ALPHA(.05)
/WSDESIGN=factor1.
Descriptive Statistics
Mean
Std. Deviation
N
v1
40.50
11.008
10
v2
42.40
11.806
10
v3
43.00
11.126
10
v4
36.80
10.390
10
Multivariate Testsa
Effect
Value
F
Hypothesis df
Error df
Sig.
factor1
Pillai\'s Trace
.768
7.725b
3.000
7.000
.013
Wilks\' Lambda
.232
7.725b
3.000
7.000
.013
Hotelling\'s Trace
3.311
7.725b
3.000
7.000
.013
Roy\'s Largest Root
3.311
7.725b
3.000
7.000
.013
a. Design: Intercept
Within Subjects Design: factor1
b. Exact statistic
Mauchly\'s Test of Sphericitya
Measure: MEASURE_1
Within Subjects Effect
Mauchly\'s W
Approx. Chi-Square
df
Sig.
Epsilonb
Greenhouse-Geisser
Huynh-Feldt
Lower-bound
factor1
.509
5.209
5
.394
.688
.896
.333
Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed
dependent variables is proportional to an identity matrix.
a. Design: Intercept
Within Subjects Design: factor1
b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected
tests are displayed in the Tests of Within-Subjects Effects table.
Tests of Within-Subjects Effects
Measure: MEASURE_1
Source
Type III Sum of Squares
df
Mean Square
F
Sig.
factor1
Sphericity Assumed
234.275
3
78.092
15.916
.000
Greenhouse-Geisser
234.275
2.064
113.520
15.916
.000
Huynh-Feldt
234.275
2.687
87.190
15.916
.000
Lower-bound
234.275
1.000
234.275
15.916
.003
Error(factor1)
Sphericity Assumed
132.475
27
4.906
Greenhouse-Geisser
132.475
18.574
7.132
Huynh-Feldt
132.475
24.182
5.478
Lower-bound
132.475
9.000
14.719
Tests of Within-Subjects Contrasts
Measure: MEASURE_1
Source
factor1
Type III Sum of Squares
df
Mean Square
F
Sig.
factor1
Linear
55.125
1
55.125
20.066
.002
Quadratic
164.025
1
164.025
18.401
.002
Cubic
15.125
1
15.125
4.946
.053
Error(factor1)
Linear
24.725
9
2.747
Quadratic
80.225
9
8.914
Cubic
27.525
9
3.058
Tests of Between-Subjects Effects
Measure: MEASURE_1
Transformed Variable: Average
Source
Type III Sum of Squares
df
Mean Square
F
Sig.
Intercept
66178.225
1
66178.225
138.576
.000
Error
4298.025
9
477.558
Estimated Marginal Means
factor1
Estimates
Measure: MEASURE_1
factor1
Mean
Std. Error
95% Confidence Interval
Lower Bound
Upper Bound
1
40.500
3.481
32.626
48..
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- 1. what is the prediction equation for the median income and incarceration rate...and the prediction equation for the total crime and incarceration rate? Incarceration Rate Median Violent Crime Prop crime Total crime per 100,00 Adults Income Rate Rate Rate 2013 2010-2013 2012 2012 2012 State Alabama 910 43330 449.9 3502.2 3952.1 Arkansas 1010 40877 469.1 3660.1 4129.2 Connecticut 620 67807 283 2140
- 3. 3071.8 Nebraska 600 55107 259.4 2754.9 3014.3 New Jersey 540 65321 290.2 2047.3 2337.5 North Carolina 730 44254 353.4 3369.5 3722.9 Oklahoma 740 47282 469.3 3401 3870.3 Rhode Island 400 55158 252.4 2572.3 2824.7 Tennessee 960 42785 643.6 3371.4
- 4. 4015 Vermont 410 56175 142.6 2398.7 2541.3 West Virginia 660 43361 316.3 2364.9 2681.2 Mean 748.82 50448.00 355.96 2801.11 3157.07 SD 273.52 9056.11 130.62 601.63 703.41 Incarceration Rate Median Violent Crime Prop crime Total crime per 100,00 Adults Income Rate Rate Rate 2013
- 8. 748.82 50448.00 355.96 2801.11 3157.07 SD 273.52 9056.11 130.62 601.63 703.41 Solution what is the prediction equation for the median income and incarceration rate...and the prediction equation for the total crime and incarceration rate? Regression Analysis r² 0.536 n 17 r -0.732 k 1 Std. Error 192.463 Dep. Var. Incarceration Rate ANOVA table Source SS df MS F p-value
- 9. Regression 641,346.1758 1 641,346.1758 17.31 .0008 Residual 555,630.2948 15 37,042.0197 Total 1,196,976.4706 16 Regression output confidence interval variables coefficients std. error t (df=15) p-value 95% lower 95% upper Intercept 1,864.1151 272.0681 6.852 5.49E-06 1,284.2157 2,444.0145 Median income -0.0221 0.0053 -4.161 .0008 -0.0334 -0.0108
- 10. Incarceration Rate =1,864.1151 - 0.0221*median income Regression Analysis r² 0.491 n 17 r 0.701 k 1 Std. Error 201.486 Dep. Var. Incarceration Rate ANOVA table Source SS df MS F p-value Regression 588,028.9910 1 588,028.9910 14.48 .0017 Residual 608,947.4796 15 40,596.4986 Total 1,196,976.4706 16 Regression output confidence interval
- 11. variables coefficients std. error t (df=15) p-value 95% lower 95% upper Intercept -111.6094 231.3010 -0.483 .6364 -604.6159 381.3970 Total crime 0.2725 0.0716 3.806 .0017 0.1199 0.4252 Incarceration Rate = -111.6094+0.2725* Total crime Regression Analysis r² 0.536 n 17 r -0.732 k 1 Std. Error 192.463 Dep. Var. Incarceration Rate ANOVA table
- 12. Source SS df MS F p-value Regression 641,346.1758 1 641,346.1758 17.31 .0008 Residual 555,630.2948 15 37,042.0197 Total 1,196,976.4706 16 Regression output confidence interval variables coefficients std. error t (df=15) p-value 95% lower 95% upper Intercept 1,864.1151 272.0681 6.852 5.49E-06 1,284.2157 2,444.0145 Median income