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jeyben [28]
3 years ago
8

For each of the following claims, fill in the blanks in the provided hypotheses with inequalities. If your answer is "less than,

" type: < If your answer is "greater than," type: > If your answer is "less than or equal to," type: <= If your answer is "greater than or equal to," type: >= If your answer is "equal to," type: = If your answer is "not equal to," type: /= (a) Theory to be tested: The average number of children per household in the United States is less than 3. H0: μ 3 and Ha: μ 3 (b) Theory to be tested: The average weight of TAMU students is greater than 110 pounds. H0: μ 110 and Ha: μ 110 (c) Theory to be tested: The average diameter of Saguaro cacti is not 2.1 feet. H0: μ 2.1 and Ha: μ 2.1 (d) Theory to be tested: In Belarus (former member of the Soviet Union, located near Chernobyl), the average life span is under 70 years. H0: μ 70 and Ha: μ 70
Mathematics
1 answer:
Leni [432]3 years ago
3 0

Answer:

a) H0: μ=>3; Ha: μ<3

b) H0: μ<=110; Ha: μ>110

c) H0: μ=2.1; Ha: μ/=2.1

d) H0: μ=>70; Ha: μ<70

Step-by-step explanation:

The claims are usually expressed in the alternative hypothesis, for which we evaluate the evidence, usually sample statistics.

(a) Theory to be tested: The average number of children per household in the United States is less than 3.

In this case, the claim is that the mean is below 3 children per household. The null hypothesis will state then that the average number of children is 3 or more.

H0: μ=>3; Ha: μ<3

(b) Theory to be tested: The average weight of TAMU students is greater than 110 pounds.

The alternative hypothesis represents the claim that the mean is greater than 110. The null hypothesis, then will state that the mean is equal or less than 110.

H0: μ<=110; Ha: μ>110

(c) Theory to be tested: The average diameter of Saguaro cacti is not 2.1 feet.

In this case, the alternative hypothesis is represented by an unequal sign (the average diameter can be higher or lower). Then, the null hypothesis will state that the average diameter is equal to 2.1.

H0: μ=2.1; Ha: μ/=2.1

(d) Theory to be tested: In Belarus (former member of the Soviet Union, located near Chernobyl), the average life span is under 70 years.

The alternative hypothesis states that the average life span is under 70 years, so the null hypothesis will state that it is equal or more than 70 years.

H0: μ=>70; Ha: μ<70

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Step-by-step explanation:

Information about the holiday:

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At the start of the football game there were 640 fans in the stadium. It begun to rain during halftime, so 10% of the fans went
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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman
Yanka [14]

Answer:

W\approx 8.72 and L\approx 15.57.

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})

\text{Area}=W\cdot L+\frac{\pi}{8}W^2

Now, we will solve for L is terms W from perimeter equation as:

L=38-(W+\frac{\pi }{2}W)

Substitute this value in area equation:

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2

Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

A'=38-2W-\pi W+\frac{\pi}{4}W    

A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

W=8.723210

W\approx 8.72

Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

L\approx 15.57

Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

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Answer:

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Step-by-step explanation:

Hope this helps!

7 0
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