How to do this problem

Is the set of positive intergers the same as nonnegative integers?

algol13

No there totally differnt

8 0

3 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago

One uranium atom has a mass of 3.95 x 10^-22 grams.

creativ13 [48]

Remember
(ab)/(cd)=(a/c)(b/d)
and
$\frac{x^n}{x^m}=x^{n-m}$

1kg=1000g=1*10^3g
hmm, I estimate, 4 of them

x=number of attoms
mass of 1 times x=1*10^3
3.95*10^-22 times x=1*10^3
divide both sides by 3.95*10^-22
x= $\frac{1*10^3}{3.95*10^{-22}}= (\frac{1}{3.95} )( \frac{10^{3}}{10^{-22}} )$ = $0.25316*10^{25}$ = $2.5316*10^{24}$
a million has 9 zeroes
so 24 zeroes is alot that is a septillion, so there are 2.5316 septillion uranium atoms in 1kg

a. 4
b. underestimate

5 0

3 years ago

Eric made a drawing of his rectangular bedroom with the scale 1 inch=3 feet. The drawing was 5 inches long 4 inches wide. What w

kolbaska11 [484]

Answer:

180 square feet

Step-by-step explanation:

We know that the rectangular drawing has:

5 inches long (L)
4 inches wide (W)

And we also know that:

1 inch = 3 feet

We will find how many feet long and wide is the actual room, by multiplying:

Long: 5 inches * 3 feet/inch = 15 feet

Wide: 4 inches * 3 feet/ inch = 12 feet

Now we want to know the area of the rectangle, and the area of the rectangle is:

A = L*W

A = 15 feet * 12 feet

A = 180 square feet

3 0

4 years ago

for every 5 pounds of pasta served , 15 people can be fed. how many people can be fed with 4 pounds of pasta?

Sidana [21]

Answer:

12

Step-by-step explanation:

its direct proportion. 15 divided by 5 is 3 so you must multiply 3 by 4 to get the amount of people, which is 12

7 0

3 years ago