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Svetlanka [38]
3 years ago
12

Quick need help for this question

Mathematics
1 answer:
Tasya [4]3 years ago
6 0
It’s c I did this in my class
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Please help!!!!!!!!!!​
Minchanka [31]

Answer:

The overall average is 70 dollars for all weeks together

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3 years ago
Read 2 more answers
On each of five slips of paper, you wrote the words eggs, waffles, pancakes, fruit, and muffin. Every day, you choose one slip f
Luda [366]
The answer is 1/5, or 0.2, because if you're always putting the slip back in, then there will always by 5 slips to choose from, therefore making the probability 1 out of 5. I hope this helps!
6 0
4 years ago
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What is the width of a rectangle with lengh 25in and area 375 in. 2
bonufazy [111]

Answer:

15in

Step-by-step explanation:

Area=width x length

Width=375/25 = 15in

8 0
4 years ago
Please help me with this question:)
Lunna [17]

Answer:

can you type the problem out please i will answer it i just the picture is kinda weird for me

Step-by-step explanation:

7 0
4 years ago
For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

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

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
4 years ago
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