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Nastasia [14]
3 years ago
8

Please Help! 10 points!

Mathematics
2 answers:
NeTakaya3 years ago
4 0

Answer:

D.

Step-by-step explanation:

Are the two triangles supposed to be similar? Because otherwise I don't see how you can do this. It's possible to have two different triangles of the same area but different perimeters, which means given a triangle, you could make two different triangles that have different perimeters yet both have an area 25 times greater than the original.  

If they're similar, then let b and h be the height of triangle A. then the base and height of triangle B are bx and hx for some unknown value of x, which means its area is (1/2)(bx)(hx) = (1/2)bh * x^2. To get this to be 25 times the area of A, (1/2)bh, just let x=5. This means the perimeter should be 5 times greater.

Julli [10]3 years ago
3 0

Part A)

<h3>Answer: 2.25</h3>

----------

You find this by squaring 1.5 to get (1.5)^2 = (1.5)*(1.5) = 2.25

Consider a rectangle that is 10 by 4, so its area is 40. Multiply each dimension by 1.5 and we get a 15 by 6 rectangle, so the new area is 90. Note how dividing 90 over 40 leads to 2.25, which is the proper area scale factor.

This example of using a rectangle can be extended to any other geometric shape.

===========================================

Part B)

<h3>Answer: 2</h3>

----------

Go in reverse of what happened in part A. Instead of squaring, we apply the square root. The square root of 4 is 2, so this is the scale factor between the perimeters.

Consider a rectangle that is 5 by 10, so it has area 50. Then consider another rectangle that is 10 by 20, with area 200. The jump from 50 to 200 has us multiply by 4, meaning the larger rectangle's area is four times larger.

Going from 5 to 10 is a jump of "times 2", so is the jump from 10 to 20. So each dimension has been multiplied by 2. Consequently, the old perimeter has been doubled to get the new one.

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A clothing store is offering 40% off of any sale merchandise. Chester's subtotal is $128.00. What is his total after the discoun
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3 years ago
Read 2 more answers
Consider random samples of size 86 drawn from population A with proportion 0.43 and random samples of size 60 drawn from populat
Elan Coil [88]

Answer:

a) The standard error is s = 0.071.

b) Yes, as both sample sizes are above 30.

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Samples of size 86 drawn from population A with proportion 0.43

This means that n = 86, p = 0.43. So:

s_A = \sqrt{\frac{0.43*0.57}{86}} = 0.0534

Samples of size 60 drawn from population B with proportion 0.15:

This means that n = 60, p = 0.15. So:

s_B = \sqrt{\frac{0.15*0.85}{60}} = 0.0461

(a) Find the standard error of the distribution of differences in sample proportions A - B Round your answer for the standard error to three decimal places. Standard error=

This is:

s = \sqrt{s_A^2 + s_B^2}

s = \sqrt{(0.0534)^2 + (0.0461)^2}

s = 0.071

The standard error is s = 0.071.

(b) Are the sample sizes large enough for the Central Limit Theorem. Yes or No?

Yes, as both sample sizes are above 30.

5 0
3 years ago
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