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Blababa [14]
3 years ago
12

Ben is planning to fence his rectangular garden. The area of the garden is 50 square feet, and the length of the garden is twice

the width. What is the length of the fence?
Mathematics
2 answers:
xxTIMURxx [149]3 years ago
8 0

Answer:

width of the fence = 5 feet

Length of the fence =10 feet

Step-by-step explanation:

Ben is planning to fence his rectangular garden. The area of the garden is 50 square feet, and the length of the garden is twice the width.

LEt x be the width of the fence

the length of the fence is twice the width.

Length of the garden = 2 * width = 2x

ARea of the garden = length * width

Plug in 2x for length and x for width

Area = 2x * x= 2x^2

Area is 50

50= 2x^2

Divide by 2 on both sides

25= x^2

Take square root on both sides

x= 5

So width of the fence = 5 feet

Length of the fence = 2x= 2*5= 10 feet

Assoli18 [71]3 years ago
5 0
X times 2x=50
2x^2=50
Divide by 2
X^2= 25
X=5

Length is 10 ft
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3 years ago
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To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data
nexus9112 [7]

Answer:

a) \sigma = 0.167

b) We need a sample of at least 282 young men.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

This Zscore is how many standard deviations the value of the measure X is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of \mu. So X - \mu = 0.5 must have Z = 3 and X - \mu = -0.5 must have Z = -3.

So

Z = \frac{X - \mu}{\sigma}

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(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

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s = \frac{\sigma}{\sqrt{n}}

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0.167 = \frac{2.8}{\sqrt{n}}

0.167\sqrt{n} = 2.8

\sqrt{n} = \frac{2.8}{0.167}

\sqrt{n} = 16.77

\sqrt{n}^{2} = 16.77^{2}

n = 281.23

We need a sample of at least 282 young men.

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