Hi,I have added a loop for adding values to list. Highlighted the .pdf
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The author added a for loop to a LinkedListAdd method to iteratively add integer values from 0 to the parameter numItems minus 1 to a LinkedList. The code highlights where a for loop was added to populate the list inside the loop using the loop counter variable i rather than individual add calls. A startTime variable tracks method execution time.
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Structure Hydra has a tubular body, with only one opening the mouth.pdf
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g
Solution
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g.
Quality management ensures that an organization, product or service .pdf
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M
Solution
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M.
public class Deque {private class Node {public int data;public.pdf
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..
Microsoft Excel is a spreadsheet program used to store and retrieve .pdf
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent.
Solution
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent..
Mean = xffS = Sqrt((x-x)2fn)intervalMidpoint(x)frequency.pdf
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3
Solution
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3.
Moles of Be = mass of Bemolar mass of Be= (0.33 g)(9.012 gmol).pdf
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further.
Solution
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further..
Lyme disease is a bacterial infection which is caused by bacteria ca.pdf
Customers table
CREATE TABLE CUSTOMERS(ID INT PRIMARY KEY,LASTNAME
CHAR(10),FIRSTNAME CHAR(10),ADDRESS VARCHAR(30),CURRENTBALANCE
INT,CREDIT_LIMIT INT,SALESREP_ID INT);
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'124\',\'ADAMS\',\'SALLY\',\'481 OAK LANSING MI
49224\',\'818.75\',\'1000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'256\',\'SAMUELS\',\'ANN\',\'215 PETE GRANT MI
49219\',\'21.5\',\'1500\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'311\',\'CHARLES\',\'DON\',\'48 COLLEGE IRA MI
49034\',\'825.75\',\'1000\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'315\',\'DANIELS\',\'TOM\',\'914 CHERRY KENT MI
48391\',\'770.75\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'405\',\'WILLIAMS\',\'AL\',\'519 WATSON GRANT MI
49219\',\'402.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'412\',\'ADAMS\',\'SALLY\',\'16 ELM LANSING MI
49224\',\'1817.5\',\'2000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'522\',\'NELSON\',\'MARY,\'108 PINE ADA MI
49441\',\'98.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'567\',\'DINH\',\'TRAN\',\'808 RIDGE HARPER MI
48421\',\'402.4\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'587\',\'GALVEZ\',\'MARA\',\'512 PINE ADA MI
49441\',\'114.6\',\'1000\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'622\',\'MARTIN\',\'DAN\',\' 419 CHIP GRANT MI
49219\',\'1045.75\',\'1000\',\'3\');
Salesreps table
CREATE TABLE SALESREPS(ID INT,LASTNAME CHAR(10),FIRSTNAME
CHAR(10),ADDRESS VARCHAR(20),TOTAL _COMMISSION
INT(10),COMMISSION_RATE INT(10));
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'3\',\'JONES\',\'MARY\',\'123 MAIN GRANT MI 49219\',\'2150\',\'.05\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'6\',\'SMITH\',\'WILLIAM\',\'102 RAYMOND ADA MI
49441\',\'4912.5\',\'.07\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'12\',\'DIAZ\',\'MIGUEL\',\'419 HARPER LANSING MI
49224\',\'2150\',\'.05\');
Orders table
CREATE TABLE ORDERS(ID INT FOREIGN KEY,ORDER_DATE DATE,CUSTOMER
INT(10),SHIPPING_DATE DATE);
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12489\',\'02-JUL-11\',\'124,\'22-JUL-11\');
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12491\',\'02-JUL-11\',\'311\',\'22-.
import java.util.Scanner; import java.util.Random; public clas.pdf
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg
Solution
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg.
Recommended
Structure Hydra has a tubular body, with only one opening the mouth.pdf
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g
Solution
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g.
Quality management ensures that an organization, product or service .pdf
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M
Solution
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M.
public class Deque {private class Node {public int data;public.pdf
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..
Microsoft Excel is a spreadsheet program used to store and retrieve .pdf
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent.
Solution
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent..
Mean = xffS = Sqrt((x-x)2fn)intervalMidpoint(x)frequency.pdf
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3
Solution
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3.
Moles of Be = mass of Bemolar mass of Be= (0.33 g)(9.012 gmol).pdf
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further.
Solution
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further..
Lyme disease is a bacterial infection which is caused by bacteria ca.pdf
Customers table
CREATE TABLE CUSTOMERS(ID INT PRIMARY KEY,LASTNAME
CHAR(10),FIRSTNAME CHAR(10),ADDRESS VARCHAR(30),CURRENTBALANCE
INT,CREDIT_LIMIT INT,SALESREP_ID INT);
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'124\',\'ADAMS\',\'SALLY\',\'481 OAK LANSING MI
49224\',\'818.75\',\'1000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'256\',\'SAMUELS\',\'ANN\',\'215 PETE GRANT MI
49219\',\'21.5\',\'1500\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'311\',\'CHARLES\',\'DON\',\'48 COLLEGE IRA MI
49034\',\'825.75\',\'1000\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'315\',\'DANIELS\',\'TOM\',\'914 CHERRY KENT MI
48391\',\'770.75\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'405\',\'WILLIAMS\',\'AL\',\'519 WATSON GRANT MI
49219\',\'402.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'412\',\'ADAMS\',\'SALLY\',\'16 ELM LANSING MI
49224\',\'1817.5\',\'2000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'522\',\'NELSON\',\'MARY,\'108 PINE ADA MI
49441\',\'98.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'567\',\'DINH\',\'TRAN\',\'808 RIDGE HARPER MI
48421\',\'402.4\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'587\',\'GALVEZ\',\'MARA\',\'512 PINE ADA MI
49441\',\'114.6\',\'1000\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'622\',\'MARTIN\',\'DAN\',\' 419 CHIP GRANT MI
49219\',\'1045.75\',\'1000\',\'3\');
Salesreps table
CREATE TABLE SALESREPS(ID INT,LASTNAME CHAR(10),FIRSTNAME
CHAR(10),ADDRESS VARCHAR(20),TOTAL _COMMISSION
INT(10),COMMISSION_RATE INT(10));
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'3\',\'JONES\',\'MARY\',\'123 MAIN GRANT MI 49219\',\'2150\',\'.05\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'6\',\'SMITH\',\'WILLIAM\',\'102 RAYMOND ADA MI
49441\',\'4912.5\',\'.07\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'12\',\'DIAZ\',\'MIGUEL\',\'419 HARPER LANSING MI
49224\',\'2150\',\'.05\');
Orders table
CREATE TABLE ORDERS(ID INT FOREIGN KEY,ORDER_DATE DATE,CUSTOMER
INT(10),SHIPPING_DATE DATE);
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12489\',\'02-JUL-11\',\'124,\'22-JUL-11\');
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12491\',\'02-JUL-11\',\'311\',\'22-.
import java.util.Scanner; import java.util.Random; public clas.pdf
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg
Solution
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg.
above molecule has 12 signlas. .pdf
above molecule has 12 signlas.
Solution
above molecule has 12 signlas..
D. ICANNThe Internet Assigned Numbers Authority (IANA) is a depart.pdf
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y
Solution
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y.
Cracking of petrolium In petroleum geology and chemistry, cracki.pdf
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9
Solution
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9.
can you provide any aditional information apart from thisSolutio.pdf
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male
Solution
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male.
brasstinSolutionbrasstin.pdf
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6
Solution
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6.
AnswerMajority of Americans have faster advancing into above 85 y.pdf
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i
Solution
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i.
AnswerA compilation error is generated The method f(int) is und.pdf
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-)
Solution
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-).
a) There is a cluster of SNP’s with similar p value in the intron .pdf
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s)
Solution
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s).
A HRIS, which is otherwise called a human resource information syste.pdf
/*
* Sorter class contains static methods for bubble sort,
* insertion sort,selection sort and quick sort.
*
* */
//Sorter.java
public class Sorter
{
//Bubble sort
public static int[] bubbleSort(int list[])
{
for (int c = 0; c < ( list.length - 1 ); c++)
{
for (int d = 0; d < list.length - c - 1; d++)
{
if (list[d]>list[d+1])
/* For ascending order*/
{
int temp= list[d];
list[d]=list[d+1];
list[d+1]=temp;
}
}
}
return list;
}
//Selection sort
public static int[] selectionSort(int[] list)
{
for (int i = 0; i < list.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < list.length; j++)
if (list[j] 0 ; j--)
{
if(list[j] pivot)
{
j--;
}
if (i <= j) {
exchangeNumbers(array,i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort()
if (lowerIndex < j)
quickSort(array,lowerIndex, j);
if (i < higherIndex)
quickSort(array,i, higherIndex);
}
//exchange elements at index i and j
private static void exchangeNumbers(int array[],int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
----------------------------------------------------------------------------------------------------------------
/**
* The java program Driver that calls the methods
* of Sorter class to sort 8 element array for bubble sort,
* insertion and selection sort.
* Call quicksort for 16 element array and print
* values before and after sorting .
* */
//Driver.java
public class Driver
{
public static void main(String[] args)
{
int arr1[]={1,7,3,2,0,5,0,8};
int arr2[]={1,7,3,2,0,5,0,8};
int arr3[]={1,7,3,2,0,5,0,8};
System.out.println(\"Before bubble sort\");
for (int i = 0; i < arr1.length; i++)
System.out.printf(\"%5d\",arr1[i]);
int bsort[]=Sorter.bubbleSort(arr1);
System.out.println(\"\ After bubble sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",bsort[i]);
System.out.println(\"\ Before insertion sort\");
for (int i = 0; i < arr2.length; i++)
System.out.printf(\"%5d\",arr2[i]);
int isort[]=Sorter.insertionSort(arr2);
System.out.println(\"\ After insertion sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",isort[i]);
System.out.println(\"\ Before selection sort\");
for (int i = 0; i < arr3.length; i++)
System.out.printf(\"%5d\",arr3[i]);
int ssort[]=Sorter.insertionSort(arr3);
System.out.println(\"\ After selection sort\");
for (int i = 0; i < ssort.length; i++)
System.out.printf(\"%5d\",ssort[i]);
int arr4[]={22,36,6,79,26,45,75,13,31,62,27,76,33,16,62,47};
System.out.println(\"\ Before quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
Sorter.quickSort(arr4, 0, arr4.length-1);
System.out.println(\"\ After quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
}
}
----------------------------------------------------------------------------------------------------------------
Sample output:
Before bubble sort
1 7 3 2 0 5 0 8
After bubble sort
0 0 1 2 3 5 7 8
Before insertion sort
1 7 3 2 0 5 0 8
After i.
4 (110000)^3 = 4 s^2 0.01s = 10^-5 MoleslitSolution.pdf
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles.
Solution
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles..
2009.88Solution2009.88.pdf
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms
Solution
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms.
1. It will be absorbed by plant. Or remain in soil that .pdf
mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% =
50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g
Solution
mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% =
50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g.
It is not that simple because TM cmplxs often have two or three band.pdf
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles
Solution
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles.
Xe (I think) because it is a bigger molecule the .pdf
Hello. Recently there has been a change in Cramster Rules (I hope you have read
the new Terms and Conditions). Cramster now bans users for giving answers directly. We have
to send the answers to your inbox after you rate us. That is because people are copying from
older threads which have public answers. Hence we can only send the answer after you rate. So
please me rate a Lifesaver and I\'ll send the solution to your inbox or email. You\'ve got no other
option to get the answer, because no one can give you the answer over here. Even you might be
banned for rating a user that gave the answer directly on your question. You need not worry as I
have the solution ready in my notebook. Hope you rate me :)
Solution
Hello. Recently there has been a change in Cramster Rules (I hope you have read
the new Terms and Conditions). Cramster now bans users for giving answers directly. We have
to send the answers to your inbox after you rate us. That is because people are copying from
older threads which have public answers. Hence we can only send the answer after you rate. So
please me rate a Lifesaver and I\'ll send the solution to your inbox or email. You\'ve got no other
option to get the answer, because no one can give you the answer over here. Even you might be
banned for rating a user that gave the answer directly on your question. You need not worry as I
have the solution ready in my notebook. Hope you rate me :).
Two of the above .pdf
first off, you dont\' have hydrogen on the left side of the equation, when something
is reduced it GAINS electrons, or gets \" more negative \" in its charge. on the left side, iodine
has a charge of + 5 when bonded with oxygen, and -1 alone on the right side, iodine has a
charge of 0 zero, oxygen\'s charge remains the same so, Iodine goes from a +5 charge to a zero
charge, getting more negative, which is reduced.
Solution
first off, you dont\' have hydrogen on the left side of the equation, when something
is reduced it GAINS electrons, or gets \" more negative \" in its charge. on the left side, iodine
has a charge of + 5 when bonded with oxygen, and -1 alone on the right side, iodine has a
charge of 0 zero, oxygen\'s charge remains the same so, Iodine goes from a +5 charge to a zero
charge, getting more negative, which is reduced..
Diffrence between totipotent and pluripotent cells are Totipotent.pdf
a ) s = 7.6/4 = 1.9
b ) P(X > 5) = P(Z > (5 - 8.2)/1.9) = P(Z > -1.68 ) = 0.9535
c ) P(X > 10) = P(Z < (10 - 8.2)/1.9) = P(Z < 0.95 ) = 0.8289
d ) guarentee = 8.2 - 1.405*1.9 = 5.5
Solution
a ) s = 7.6/4 = 1.9
b ) P(X > 5) = P(Z > (5 - 8.2)/1.9) = P(Z > -1.68 ) = 0.9535
c ) P(X > 10) = P(Z < (10 - 8.2)/1.9) = P(Z < 0.95 ) = 0.8289
d ) guarentee = 8.2 - 1.405*1.9 = 5.5.
The number of mols of NaF = 0.15 molL0.045 L .pdf
deltaHrxns = -200 - (-100) = -100 kJ
Solution
deltaHrxns = -200 - (-100) = -100 kJ.
How to Build a Module in Odoo 17 Using the Scaffold Method
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar KanvariaDelivering Micro-Credentials in Technical and Vocational Education and Training
Explore how micro-credentials are transforming Technical and Vocational Education and Training (TVET) with this comprehensive slide deck. Discover what micro-credentials are, their importance in TVET, the advantages they offer, and the insights from industry experts. Additionally, learn about the top software applications available for creating and managing micro-credentials. This presentation also includes valuable resources and a discussion on the future of these specialised certifications.
For more detailed information on delivering micro-credentials in TVET, visit this https://tvettrainer.com/delivering-micro-credentials-in-tvet/
The Challenger.pdf DNHS Official Publication
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
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.
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above molecule has 12 signlas. .pdf
above molecule has 12 signlas.
Solution
above molecule has 12 signlas..
D. ICANNThe Internet Assigned Numbers Authority (IANA) is a depart.pdf
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y
Solution
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y.
Cracking of petrolium In petroleum geology and chemistry, cracki.pdf
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9
Solution
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9.
can you provide any aditional information apart from thisSolutio.pdf
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male
Solution
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male.
brasstinSolutionbrasstin.pdf
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6
Solution
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6.
AnswerMajority of Americans have faster advancing into above 85 y.pdf
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i
Solution
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i.
AnswerA compilation error is generated The method f(int) is und.pdf
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-)
Solution
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-).
a) There is a cluster of SNP’s with similar p value in the intron .pdf
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s)
Solution
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s).
A HRIS, which is otherwise called a human resource information syste.pdf
/*
* Sorter class contains static methods for bubble sort,
* insertion sort,selection sort and quick sort.
*
* */
//Sorter.java
public class Sorter
{
//Bubble sort
public static int[] bubbleSort(int list[])
{
for (int c = 0; c < ( list.length - 1 ); c++)
{
for (int d = 0; d < list.length - c - 1; d++)
{
if (list[d]>list[d+1])
/* For ascending order*/
{
int temp= list[d];
list[d]=list[d+1];
list[d+1]=temp;
}
}
}
return list;
}
//Selection sort
public static int[] selectionSort(int[] list)
{
for (int i = 0; i < list.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < list.length; j++)
if (list[j] 0 ; j--)
{
if(list[j] pivot)
{
j--;
}
if (i <= j) {
exchangeNumbers(array,i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort()
if (lowerIndex < j)
quickSort(array,lowerIndex, j);
if (i < higherIndex)
quickSort(array,i, higherIndex);
}
//exchange elements at index i and j
private static void exchangeNumbers(int array[],int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
----------------------------------------------------------------------------------------------------------------
/**
* The java program Driver that calls the methods
* of Sorter class to sort 8 element array for bubble sort,
* insertion and selection sort.
* Call quicksort for 16 element array and print
* values before and after sorting .
* */
//Driver.java
public class Driver
{
public static void main(String[] args)
{
int arr1[]={1,7,3,2,0,5,0,8};
int arr2[]={1,7,3,2,0,5,0,8};
int arr3[]={1,7,3,2,0,5,0,8};
System.out.println(\"Before bubble sort\");
for (int i = 0; i < arr1.length; i++)
System.out.printf(\"%5d\",arr1[i]);
int bsort[]=Sorter.bubbleSort(arr1);
System.out.println(\"\ After bubble sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",bsort[i]);
System.out.println(\"\ Before insertion sort\");
for (int i = 0; i < arr2.length; i++)
System.out.printf(\"%5d\",arr2[i]);
int isort[]=Sorter.insertionSort(arr2);
System.out.println(\"\ After insertion sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",isort[i]);
System.out.println(\"\ Before selection sort\");
for (int i = 0; i < arr3.length; i++)
System.out.printf(\"%5d\",arr3[i]);
int ssort[]=Sorter.insertionSort(arr3);
System.out.println(\"\ After selection sort\");
for (int i = 0; i < ssort.length; i++)
System.out.printf(\"%5d\",ssort[i]);
int arr4[]={22,36,6,79,26,45,75,13,31,62,27,76,33,16,62,47};
System.out.println(\"\ Before quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
Sorter.quickSort(arr4, 0, arr4.length-1);
System.out.println(\"\ After quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
}
}
----------------------------------------------------------------------------------------------------------------
Sample output:
Before bubble sort
1 7 3 2 0 5 0 8
After bubble sort
0 0 1 2 3 5 7 8
Before insertion sort
1 7 3 2 0 5 0 8
After i.
4 (110000)^3 = 4 s^2 0.01s = 10^-5 MoleslitSolution.pdf
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles.
Solution
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles..
2009.88Solution2009.88.pdf
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms
Solution
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms.
1. It will be absorbed by plant. Or remain in soil that .pdf
mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% =
50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g
Solution
mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% =
50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g.
It is not that simple because TM cmplxs often have two or three band.pdf
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles
Solution
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles.
Xe (I think) because it is a bigger molecule the .pdf
Hello. Recently there has been a change in Cramster Rules (I hope you have read
the new Terms and Conditions). Cramster now bans users for giving answers directly. We have
to send the answers to your inbox after you rate us. That is because people are copying from
older threads which have public answers. Hence we can only send the answer after you rate. So
please me rate a Lifesaver and I\'ll send the solution to your inbox or email. You\'ve got no other
option to get the answer, because no one can give you the answer over here. Even you might be
banned for rating a user that gave the answer directly on your question. You need not worry as I
have the solution ready in my notebook. Hope you rate me :)
Solution
Hello. Recently there has been a change in Cramster Rules (I hope you have read
the new Terms and Conditions). Cramster now bans users for giving answers directly. We have
to send the answers to your inbox after you rate us. That is because people are copying from
older threads which have public answers. Hence we can only send the answer after you rate. So
please me rate a Lifesaver and I\'ll send the solution to your inbox or email. You\'ve got no other
option to get the answer, because no one can give you the answer over here. Even you might be
banned for rating a user that gave the answer directly on your question. You need not worry as I
have the solution ready in my notebook. Hope you rate me :).
Two of the above .pdf
first off, you dont\' have hydrogen on the left side of the equation, when something
is reduced it GAINS electrons, or gets \" more negative \" in its charge. on the left side, iodine
has a charge of + 5 when bonded with oxygen, and -1 alone on the right side, iodine has a
charge of 0 zero, oxygen\'s charge remains the same so, Iodine goes from a +5 charge to a zero
charge, getting more negative, which is reduced.
Solution
first off, you dont\' have hydrogen on the left side of the equation, when something
is reduced it GAINS electrons, or gets \" more negative \" in its charge. on the left side, iodine
has a charge of + 5 when bonded with oxygen, and -1 alone on the right side, iodine has a
charge of 0 zero, oxygen\'s charge remains the same so, Iodine goes from a +5 charge to a zero
charge, getting more negative, which is reduced..
Diffrence between totipotent and pluripotent cells are Totipotent.pdf
a ) s = 7.6/4 = 1.9
b ) P(X > 5) = P(Z > (5 - 8.2)/1.9) = P(Z > -1.68 ) = 0.9535
c ) P(X > 10) = P(Z < (10 - 8.2)/1.9) = P(Z < 0.95 ) = 0.8289
d ) guarentee = 8.2 - 1.405*1.9 = 5.5
Solution
a ) s = 7.6/4 = 1.9
b ) P(X > 5) = P(Z > (5 - 8.2)/1.9) = P(Z > -1.68 ) = 0.9535
c ) P(X > 10) = P(Z < (10 - 8.2)/1.9) = P(Z < 0.95 ) = 0.8289
d ) guarentee = 8.2 - 1.405*1.9 = 5.5.
The number of mols of NaF = 0.15 molL0.045 L .pdf
deltaHrxns = -200 - (-100) = -100 kJ
Solution
deltaHrxns = -200 - (-100) = -100 kJ.
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