Most bacteria that cause diseases in humans have a temperature optim.pdf

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It is believed that the number of drivers who are ticketed in an hour for speeding on a particular stretch of highway is a Poisson distribution with a mean of 3.0 per hour. A random sample of 100 hours is selected with the following results: These data are included in the Final Exam Excel Files under the tabTicketsFrequency051922031842051564768192 Solution For testing whether it is a Poisson find mean and std dev. As mean and variance are considerably difference this data does not fit a Poisson modelTicketsFrequencytickets*freqtickets^2*freq0500199812204080031854972420801600515 7511256424967642252818892918Total1003414952Mean3.41Variance37.8919.

Most bacteria that cause diseases in humans have a temperature optimum of approximately?
Why?
Solution
The bacteria which cause foodborne illness in humans grow best at body temperature98.6o F
,which are called as Mesophilic bacteria.
Bacteria grow best within certain temperature ranges.Those temperatures are very specific to the
type of Micro organisms.
For Mesophilic bacteria Optimum temperature is 68-1130 F.,It s growth range 68-1100 F.

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It is hypothesized that when homing pigeons are disoriented in a certain manner, they will exhibit no preference for any direction of flight after takeoff (so that the direction X should be uniformly distributed on the interval from 0o to 360o). To test this, 120 pigeons are disoriented, let loose, and the direction of flight of each is recorded; the resulting data follows. Use the chi- squared test at level 0.10 to see whether the data supports the hypothesis. Direction 10-45° 45- Solution We expect 120/8 = 15 on each direction. Doing an observed/expected value table, O E (O - E)^2/E 11 15 1.066666667 17 15 0.266666667 17 15 0.266666667 15 15 0 13 15 0.266666667 19 15 1.066666667 17 15 0.266666667 11 15 1.066666667 0 0 0 0 Using chi^2 = Sum[(O - E)^2/E], chi^2 = 4.266666667 = 4.27 [ANSWER].

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4. Why are women more prone to urinary tract infections? Solution The main reason for the women more frequently get urinary tract infections is because their urethra a tube that carries urine out of the body is smaller than men\'s. Thus, the bacteria that cause the infections have easily travel up to the bladder. Risk factors specific to women for UTIs include: 1. Female anatomy, 2. Sexual activity, 3. Certain types of birth control, 4. Menopause, 5. Urinary tract abnormalities, 6. A suppressed immune system, 7. Catheter use, 8. A recent urinary procedure etc....

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13 What type of joint is present at ll, the Hinge Ball and Sliding Fixed 14) What is the name given to the tissue that transmits water and dissolved minerals from the root of a plant up towards the O c. Phloem Xylem Parenchyma Cambium 15) What tissues provides support to hold up the stem of a plant? B.O c. Phloem Xylem Parenchyman Cambium 16 What is the process that moves water through a plant? Guttation Replication Transpiration Germination 170 If a plant is described as xerophytic, what conditions will it live and grow in? A. O B. O O D. Dry Salty Wet Cold conditions soils Soils 18 What name is given to the type of cell that carries oxygen in blood? Lymphocyte Leucocyte Erythrocyte Phagocyte 19) What is the red pigment present in cells that carry oxygen? Solution Q. 13) - A. Q. 14) - B. Q 15) - C. Q 16) - C. Q 17) - D. Q 18) - C. Q 19) - A. Q 20) - D. Q 21) - B. Q 22) - C. Q 23) - A. Q 24) - B. Q 25) - A. Q 26) - C. Q 27) - C. Q 28) - C. Q 29) - D. Q 30) - C..

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Question 1 Not yet answered Marked out of 10.00 A preference share pays a $15 dividend annually. If money is worth 11 % pa, what should an investor consider paying for this stock? Answer Flag question Next page Week 2 T218 (3) pptx FIN201 Week 3 Ann....doc Solution Answer: For example, if the face value of the share = 100 Dividend per share = 15 He needs to have 11% on the value of his money = 11/100 Hence money paid for a share = 15 *100 11 = $136.36 per share..

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One of the major areas that gives rise to off balance sheet financing is leasing. a) Explain the impact of leasing transaction on the financial statements of the lessee according to the current accounting standard for leasing (IAS 17). Support your answer with an appropriate numerical example. b) At the beginning of 2016, IASB issued a new standard (IFRS 16) on leasing accounting. Explain the reasons behind this decision and the impact on companies’ financial statements. Solution a. According to IAS 17 the financial statements of lessee will be impacted during initial accounting and during subsequent accounting. The current accounting standard for leasing requires the lessee to capitalize the finance leased asset and at the same time set up a lease liability for the value of the asset recognized. The balance sheet will be affected here. Subsequent accounting will affect the financial statements as the lessee will be required to account for depreciation, lease rental and interest. These will have an impact on lessee’s income statement and statement of financial position. Numerical example: Suppose that in April 1, 2017, XYZ Co. enters into a lease with regards to machinery that has an estimated life of four years. Lease period is also four years and the machinery will be returned to the leasing company after this period. Annual rental amount is $5,000 and this amount will be paid in arrears starting from March 31, 2018. Residual value will be nil and fair value is $15,000. Finance cost is 15% per annum. Initial accounting impact: Subsequent accounting impact: Finance cost = 15% of 15,000 = $2,250 b. IFRS 16 was introduced with an intention to ensure that all lessee leases should be reported on the balance sheet and not as off balance sheet items. The main purpose was to eliminate all guess work that is involved in computing a company’s lease obligations. IFRS 16 intends to develop and introduce more transparency with regards to lease assets and liabilities. Lessee will be required to recognize the asset for the right to use the leased item. Liability would be recognized to the effect of present value of future lease payments. Lessor accounting will remain similar to current practice in terms of lease classification test and in terms of finance leases and operating leases. Sale and leaseback will no longer be valid as an off balance sheet financing structure.ParticularsDebitCreditProperty, plant and equipment15,000.00Finance lease obligations15,000.00.

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Is 53 equivalent to 2 (mod 3), 1(mod 4), 3 (mod 5)? We\'re trying to explain the Chinese remainder theorem in simple words. We got the equations x=60k+53 for any k=0,1,2,3... Is 53 equivalent to 2 (mod 3), 1(mod 4), 3 (mod 5)? We\'re trying to explain the Chinese remainder theorem in simple words. We got the equations x=60k+53 for any k=0,1,2,3... Solution we got x=60k +53 for k=0,1,2,3........... since for k=0, x=60(0)+53 x=0+53 x=53 so 53 equivalent to 2 (mod 3), 1(mod 4), 3 (mod 5).

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PLEASE ANSWER MY QUESTIONS PROPERLY WITH SOLUTION-THANK YOU You have just leased a car that has monthly payments of $305 for the next 4 years with the first payment due today. If the APR is 5.4 percent compounded monthly, what is the value of the payments today? O $13,140.36 $13,552.85 $11.999.54 O $13,199.49 $12,858.49 Solution Answer to Question 1: Monthly Payment = $305 Period = 4 years or 48 months Annual Interest Rate = 5.4% Monthly Interest Rate = 0.45% Present Value of Payments = $305 + $305/1.0045 + $305/1.0045^2 + ... + $305/1.0045^47 Present Value of Payments = $305 * 1.0045 * (1 - (1/1.0045)^48) / 0.0045 Present Value of Payments = $305 * 43.2770 Present Value of Payments = $13,199.49 Answer to Question 2: Quarterly Payment = $1,300 Period = 21 years or 84 quarters Quarterly Interest Rate = 0.93% Present Value of Payments = $1,300/1.0093 + $1,300/1.0093^2 + $1,300/1.0093^3 + ... + $1,300/1.0093^84 Present Value of Payments = $1,300 * (1 - (1/1.0093)^84) / 0.0093 Present Value of Payments = $1,300 * 58.1170 Present Value of Payments = $75,552.06 Answer to Question 3: Decrease in Accounts Receivable = $26,650 - $24,440 Decrease in Accounts Receivable = $2,210 Increase in Accounts Payable = $40,560 - $39,170 Increase in Accounts Payable = $1,390 Increase in Inventory = $22,820 - $18,930 Increase in Inventory = $3,890 Use of Cash = Increase in Inventory - Decrease in Accounts Receivable - Increase in Accounts Payable Use of Cash = $3,890 - $2,210 - $1,390 Use of Cash = $290 Answer to Question 4: Cash Cycle = Operating Cycle - Accounts Payable Period Cash Cycle = 54.2 days - 37.1 days Cash Cycle = 17.1 days Operating Cycle = Inventory Period + Accounts Receivable Period Operating Cycle = 22.5 days + 31.7 days Operating Cycle = 54.2 days.

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Question 10 (1.2 points) d What helps explain the popularity of mutua.pdf

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Let X be the mean of a random sample of size n = 12 from the uniform distribution on the interval (0, 1). Approximate P(1/2 X 2/3). Solution as its given that n =12 not infinity we can get the sidtribution of x , the sampling distribution of X N(µ, ²/n) as n 12. by using this we can the sampling distribution N(µ, ²/n) = N(1/2, (1/12)/12) = N(1/2, 1/144) now what we need to find is that x lines in range of 1/2 and 2/3 and we can do this by by dividing the variance and subtract the mean Let Z = (X-bar - µ) / (²/n) = (X-bar - 1/2) / (1/144) = 144(X-bar - 1/2) let it be z variable , which is standard normal and variance = 1 and mean =0 now we have to just it in probability expression and look from the table P(1/2 < X-bar < 2/3) = P{144(1/2 - 1/2) < Z < 144(2/3 - 1/2)} = P{0 < Z < 24} = 0.50000 by normal table.

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1. Substance’s Effect: Increases the number of free hydrogen ions or protons (H+) in the intermembrane space of the mitochondria (between the outer and inner membranes). Facilitator: __________________________________ Explanation for effect: ______________ 2. Substance’s Effect: Impedes the binding of NAD+ to electrons. Facilitator: __________________________________ 3. Substance’s Effect: Makes the lipid bilayer of the inner mitochondrial membrane highly permeable to (H+) protons. Facilitator: __________________________________ 4. Substance’s Effect: Partially blocks the H+/pyruvate symporter, interfering with the transport of pyruvate into the mitochondria. Facilitator: __________________________________ Solution At the inner mitochondria membrane, a high energy electron is passed along an electron transport chain. The energy released pumps hydrogen out of the matrix space. The gradient create by this drives hydrogen back through the membrane, through ATP synthase. As it happens the enzymatic activity of ATP synthase synthesiszes ATP from ADP. In livingcells 2,4- Dinitrophenol acts as a proton carrier, an that can shuttle protons(hydrogen ions) across biological membranes. It defeats the proton gradient across mitochondria, collapsing the proton motive force that the cell uses to produce most of itsATP chemical energy. Instead of producing ATP , the energy of the proton gradient is lost as heat- causing you to loss weight. Therefore causes weight loss. Lipid bilayer of the inner mitochondrial membrane highly leaky to protons making the inner mitochondrial membrane more leaky to h, makes the lipid bilayer of the mitochondrial membrane ,highly pereameable ,makes the membrane highly porustoprotons. This occurs in mitochondrial matrix, pyruate made in the cytosol enters through a channel(outer membrane) and then an H+/pyruvate symptorter(inner membrane) pyruvate dehydrogenase which removes CO2 from each pyruvateproducing 2H+ and 2NAD+ (which combine)..

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1. What are three functions of an operating system? 2. What are two key issues to include in contract negotiations when considering new software applications? Solution 1)Operating System is the interface between human and computer hardware, and It provides following functions >>> Resource anagement(CPU,Memory,HDD,Printers etc) >>> User Interface Establishment >>> Rinning and providing resources for the application software. 2)Business Issues , Liability Issues, Technical Issues..

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JAVA: Write a class to manipulate a string made up of a person\'s first name, middle initial, and last name. The entire name has to be created as one field and not three separate fields. The entire name must also been created as a mixture of uppercase and lowercase letters which simply is not proper. Your goal is to output the three components of the name as separate fields with proper uppercase and lowercase letters. Example: Input: First Name Initial Last Name JoHn J sMiTh Output: John J Smith Solution Names.java import java.util.Scanner; public class Names { public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.println(\"Enter First Name Initial Last Name: \"); String name = scan.nextLine(); String firstName = name.split(\"\\\\s+\")[0]; String initial = name.split(\"\\\\s+\")[1]; String lastName = name.split(\"\\\\s+\")[2]; String newFirstName = \"\", newInitial = \"\", newLastName = \"\"; for(int i=0; i.

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Hello, I need help to to add modify the program to process as many telephone numbers as the user wants and also to improve/simply my final output to fulfill this requirement - The output should display a hyphen after the first 3 digits and subsequently a hyphen (-) after four digits. My program below: import java.util.*; public class teleAlphaTwo { public static void main(String [] args) { Scanner gaz = new Scanner (System.in); int length, num; int count=0; String letters, output=\"\"; System.out.println(\"Please type the words to be converted to a phone number:\"); letters = gaz.nextLine().toLowerCase(); length = letters.length(); for (num=0; num<=length-1; num++) { switch (letters.charAt(num)) { case \'a\': case \'b\': case \'c\': output+=\"2\"; count++; break; case \'d\': case \'e\': case \'f\': output+=\"3\"; count++; break; case \'g\': case \'h\': case \'i\': output+=\"4\"; count++; break; case \'j\': case \'k\': case \'l\': output+=\"5\"; count++; break; case \'m\': case \'n\': case \'o\': output+=\"6\"; count++; break; case \'p\': case \'q\': case \'r\': case \'s\': output+=\"7\"; count++; break; case \'t\': case \'u\': case \'v\': output+=\"8\"; count++; break; case \'w\': case \'x\': case \'y\': case \'z\': output+=\"9\"; count++; break; default: output=output; break; } System.out.println(\"The phone number is \"+output.substring(0, 3)+\"-\"+output.substring(3, 7)+\"-\"+output.substring(7, 11)+\"-\"+output.substring(11, 15)); } } } Solution Java Code: import java.util.Scanner; public class TeleAlphaTwo { public static void main(String [] args) { Scanner gaz = new Scanner(System.in); int length=0; int count=0; String letters, output=\"\"; System.out.println(\"Please type the words to be converted to a phone number:\"); letters = gaz.nextLine().toLowerCase(); for(int num=0;num.

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For the purposes of this question a household corresponds to a single residential address, while household size is the number of individuals living at that address. Say you conduct a simple random sample of individuals living in Vancouver. Those sampled are asked to report their household sizes. The sample average of household size will be: A. systematically biased upwards as an estimator of the average size of Vancouver households. B. systematically biased downwards as an estimator of the average size of Vancouver households. C. sensible as an estimator of the average size of Vancouver households. Solution Note that the mean of a sampling distribution is the population mean, and the sample mean is an unbiased estimator of the population mean. Thus, C. sensible as an estimator of the average size of Vancouver households (answer).

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Find the slope of the tangent line to the ellipse x2/4+ y2/9 = 1 at the point (x,y). slope = Are there any points where the slope is not defined? (Enter them as comma-separated ordered- pairs, e.g., (1,3), (-2,5). Enter none if there are no such points.) slope is undefined at Solution y=3sqrt(1-x^2/4) dy/dx=(3/2)(-x/2)/(sqrt(1-x^2/4)) =-3x/2sqrt(4-x^2) slope is not defined where x=2,-2 slope tends to infinity (2,0),(-2,0).

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Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete parts (a) through (d) below. Click here to view page 1 of the cumulative standardized normal distribution table. LOADING... Click here to view page 2 of the cumulative standardized normal distribution table. LOADING... a. What is the probability that Z is between negative 1.551.55 and 1.881.88? The probability that Z is between negative 1.551.55 and 1.881.88 is nothing. (Round to four decimal places as needed.) Solution Normal Distribution Mean ( u ) =0 Standard Deviation ( sd )=1 Normal Distribution = Z= X- u / sd ~ N(0,1) a) To find P(a < = Z < = b) = F(b) - F(a) P(X < -1.55) = (-1.55-0)/1 = -1.55/1 = -1.55 = P ( Z <-1.55) From Standard Normal Table = 0.06057 P(X < 1.88) = (1.88-0)/1 = 1.88/1 = 1.88 = P ( Z <1.88) From Standard Normal Table = 0.96995 P(-1.55 < X < 1.88) = 0.96995-0.06057 = 0.9094.

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Comment on the 20 percent deviation from baseline cost. Is 20% reasonable? Is 20% too high or too low? Search the Internet for other examples of project oversight (good or bad). Summarize your findings. Solution yes, i searched many sites and i can say that maximum 20% is acceptable and beyond that is not acceptable to any project. ofcorse in this risk and uncertain world everything has been changing for day to day. in these conditions estimating a period of 3 or 4 years for a project is more risky and there are many chances to more deviations. but to be a professional project manager we need to consider all these and we should plan it if the variation is as possible as less, that is good to the firm..

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Explain why inhibitors that prevent plasma membrane localization of Ras have been used in clinical trials for treatment of several tumors including colon cancer. Explain why inhibitors that prevent plasma membrane localization of Ras have been used in clinical trials for treatment of several tumors including colon cancer. Solution In MAPK pathway, the binding of growth hormone to the receptor in the plasma membrane causes dimerization of the receptor (conformational change) that result in autophosphorylation of tyrosine residues located on the intracellular domains of receptors. The binding of SH2-domain (src homology) to the phosphorylated tyrosine residues stimulates the exchange of GDP bound to Ras with the GTP, now the Ras is activated. Activated Ras activates Raf, Raf activates MEK kinase, MEK kinase activates MAPK, that is followed by a kinase cascade and finally gene transcription occurs in the nucleus. In many of the cancers, constituent activation of Ras is observed, which allows continuous cell division. Thus, prevention of plasma membrane localization of Ras have been used in clinical trials for treatment of several tumors including colon cancer..

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Design a block diagram in RTL an 8-bit Data Processing Unit with the following specifications and requirements: 1- The ALU is capable of executing the following functions: a- load, store, shift-left, and shift-right b- add, add-with-carry, sub, increment c- and, or, xor, and complement d- set-a-flag, clear-a-flag, test-a-flag e- branch-if-zero, and branch-if-carry 2- Use a file register ALU architecture organization for yourdesign.3- Include the following flags in your design: C, Z, V, &N. 4- Include documentation describing your design at the register level (RTL), which should include the following: a- Hardware design for the data processing unit (which includes the bus system) that is capable of executing the required functions b- A block diagram for your processor architecture c- State all the control signals that are needed for your design (you have to justification your choices) Solution use ieee.std_logic_unsigned.all; entity alu is port( din1 : in STD_LOGIC_VECTOR(3 downto 0); din2 : in STD_LOGIC_VECTOR(3 downto 0); sel : in STD_LOGIC_VECTOR(3 downto 0); dout : out STD_LOGIC_VECTOR(3 downto 0) ); end alu; architecture alu_arc of alu is begin with sel select dout <= din1 and din2 when \"0000\", din1 or din2 when \"0001\", din1 nand din2 when \"0010\", din1 nor din2 when \"0011\", din1 xor din2 when \"0100\", din1 xnor din2 when \"0101\", not din1 when \"0110\", not din2 when \"0111\", din1 + \"0001\" when \"1000\", din2 + \"0001\" when \"1001\", din1 + din2 when \"1010\", din1 - din2 when \"1011\", din2 - \"0010\" when \"1100\", din1 - \"0010\" when \"1101\", din1 + \"0010\" when \"1110\", din2 + \"0010\" when others; end alu_arc;.

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