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A farmer wants to fence a small rectangular yard next to a barn. Fence for side parallel to the barn will cost 55 per foot and t

Mashutka [201]

Answer:

55 divided by 45= 1.222222222222 1650 divided by 1.222222= anserw

Step-by-step explanation:

6 0

3 years ago

The length of a side of a cube can be determinad

erik [133]

Answer:

4 cm

which is none of the listed answers

Step-by-step explanation:

∛64 = 4

proof:

4x4x4 = 64

7 0

3 years ago

Burger town sells cheeseburgers for $7.95 each plus an additional $1.00 for each extra toppings, t. Which of the following equat

dezoksy [38]

Answer: $7.95+$1t ?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

How do you Round 96,286 to the nearest thousand.

Varvara68 [4.7K]

Answer:

The correct answer is 96,000

Step-by-step explanation:

To round you look at the thousands place and there is a 9 then you look at the number to the right of it which is 2 and it is lower than 5 since it is lower than five it rounds down so the correct answer is 96,000.

HAVE A GOOD DAY!

8 0

3 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago