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2 years ago

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USING STRUCTURE An athletic facility is building an indoor track. The track is composed of a rectangle and two semicircles, as s

hown. A drawing shows an indoor track. It is shaped like a rectangle with a semi-circle on each of the shorter sides. The length of the rectangle is labeled x. The radius of each semi-circle is labeled r. A. Write a formula for the perimeter of the indoor track. $P=$ b. Solve the formula for $x$ . $x=$ Question 2 c. The perimeter of the track is 660 feet, and $r$ is 50 feet. Find $x$ . Round your answer to the nearest foot. $x=$ feet

1 answer:

Nesterboy [21]2 years ago

4 0

Answer:

Step-by-step explanation:

Find the digram attached.

Perimeter of the track = perimeter of the rectangle + perimeter of the 2semicircles

Perimter of a rectangle = 2(x+r) where:

x is the length

2r is the width of the rectangle = diameter of the semicircle

Perimeter of semicircle = 2πr/2 = πr

Perimeter of 2semicirle = 2πr

Perimeter of the track = 2(x+2r) + 2πr

r is the radius if the semicircle

Expand

Perimeter of the track = 2x+4r + 2πr

Perimeter of the track = 2(x+2r+πr)

b) Given P = 2(x+2r+πr), we are to make x the subject of the formula.

P = 2x+4r+2πr

P-4r-2πr = 2x

Divide both sides by 2.

(P-4r-2πr)/2 = 2x

x = (P-4r-2πr)/2

c) Given

P = 600fr

r = 50ft

x = (600-4(50)-2π(50))/2

x = (600-200-100(3.14))/2

x = 400-314/2

x = 86/2

x = 43ft

Hence the value of x to nearest foot is 43ft

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a sheet of acrylic has a mass of 95.2g. Acrylic has a density of 1.19g/cm³ What is the volume of the sheet of acrylic

professor190 [17]

Answer:

80 cm³

Step-by-step explanation:

Density = mass/volume

1.19 = 95.2/volume

Volume = 95.2/1.19

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2 years ago

Read 2 more answers

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

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iogann1982 [59]

I'm assuming this is multiple choice and you forgot to post the answers. I'll take a guess and say it probably looks something like this:
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Answer:

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Step-by-step explanation:

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Hey there!

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