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stepan [7]
3 years ago
8

Example 4: Statistical Significance A sample of 50 cans of tomatoes are tested for levels of the chemical BPA to see if there is

evidence that the mean level is greater than 100 ppb (parts per billion). Write down the hypotheses for this test: Give a possible sample mean that you think would be statistically significant: Give a possible sample mean that would definitely not be statistically significant:
Mathematics
1 answer:
Tanya [424]3 years ago
8 0

Answer:

1.

h0: μ = 100

h1: μ > 100

2.

sample mean of 150 will be statistically significant

3

sample mean of 40 will be statistically insignificant

Step-by-step explanation:

1

the null hypothesis:

H0: μ = 100

the alternate hypothesis:

H1: μ > 100

2.

a sample mean that is likely to be statistically significant is one whose value is quite bigger than 100.

a sample mean of 150 will provide strong evidence and is therefore statistically significant

3.

a sample mean that would not be statistically significant is one whose value is less than 100. a sample mean of 40 will not be statistically significant.

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Hey there! :D

When looking at the data sets, notice that they only overlap in the middle, by a few points. There is overlap, so it isn't "none", it isn't much, so it wouldn't be "high" or "moderate". It would be considered low overlap. 

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Detailed Answer:


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An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as
Paul [167]

Answer:

A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is \frac{a}{\sqrt{3}}=R.

Upon substituting our given values, we will get:

\frac{5x}{\sqrt{3}}=6r

Let us solve for r.

r=\frac{5x}{6\sqrt{3}}

\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}

We know that area of an equilateral triangle is equal to \frac{\sqrt{3}}{4}s^2, where s represents side length of triangle.

\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2

The area within circle and outside the triangle would be difference of area of circle and triangle as:

A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}

We can make a common denominator as:

A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}

A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}

Therefore, our required expression would be A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}.

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3 years ago
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