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Vladimir79 [104]
4 years ago
5

Laura's yard is in the shape of a square and a half-circle.

Mathematics
2 answers:
DanielleElmas [232]4 years ago
8 0
Pi r2

17 divide 2=8.5

Then

8.5•8.5=72.25

Then

3.14•72.25=226.865

Then

226.865 divide by 2 =113.4325m2
Lady bird [3.3K]4 years ago
8 0

Answer:

Option 2 - 402 m sq.

Step-by-step explanation:

Given : Laura's yard is in the shape of a square and a half-circle.

To find : What is the approximate area of Laura's yard?

Solution :

The side of the square is 17 m

The area of the square is A_s=s^2

A_s=17^2

A_s=289m^2

The half circle form upon a square side.

So, The diameter of the half circle is equal to the side of the square.

d=17 m

Radius of the half circle is r=\frac{17}{2}

The area of the half circle is A=\frac{1}{2} \pi r^2

Substitute the value in the formula,

A_c=\frac{1}{2}\times \frac{22}{7}\times(\frac{17}{2})^2

A_c=\frac{22\times 17\times 17}{2\times 7\times 2\times 2}

A_c=\frac{6358}{56}

A_c=113.53m^2

Total area of the Laura's yard is

A=A_s+A_c

A=289+113.53

A=402.53m^2

Approximately the area of the Laura's yard is 402 m sq.

Therefore, Option 2 is correct.

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Answer:

a) \hat p=\frac{471}{1024}=0.460

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And the margin of error is given by:

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Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

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Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

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Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491

The 99% confidence interval would be given by (0.429;0.491)

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