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2 years ago

A company makes square picture frames whose sides are integer lengths. Select the area that would satisfy this requirement

1 answer:

4 0

Area of a square is a^2...with a being the length of 1 side

A. 14 = a^2......sqrt 14 = a.....3.74...not an integer
B. 16 = a^2......sqrt 16 = a.....4....this works
C. 1.5 = a^2....sqrt 1.5 = a.....1.22....not an integer
D. 18 = a^2....sqrt 18 = a.......4.24....not an integer

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What is the answer to 446,571÷87

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Jessa works at a potato chip factory and would like to estimate the mean weight in grams of the factory's potato

Svetlanka [38]

Answer:

The smallest sample size required to obtain the desired margin of error is 44.

Step-by-step explanation:

I think there was a small typing error, we have that $\sigma = 60$ is the standard deviation of these weighs.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

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Now, find the margin of error M as such

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Which of these is the smallest approximate sample size required to obtain the desired margin of error?

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n is found when $M = 15$

$M = z*\frac{\sigma}{\sqrt{n}}$

$15 = 1.645*\frac{60}{\sqrt{n}}$

$15\sqrt{n} = 1.645*60$

Simplifying by 15

$\sqrt{n} = 1.645*4$

$(\sqrt{n})^{2} = (1.645*4)^{2}$

$n = 43.3$

Rounding up

The smallest sample size required to obtain the desired margin of error is 44.

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2 years ago

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Step 2 is the answer

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