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Vinil7 [7]
3 years ago
5

The student council at mountain view high school plans to conduct a survey to determine the student body's opinion of the new lu

nch menu. Which of the following sampling methods would be most likely to result in UNDERCOVERAGE bias?
A. Setting up a booth in the lunchroom where students can give their thoughts

B. Obtaining an alphabetical list of all students and surveying every 10th student

C. Handing out a questionnaire to seniors only

D. Handing out a survey to the first 50 students entering the cafeteria at lunch
Mathematics
1 answer:
valkas [14]3 years ago
4 0

Answer:

c

Step-by-step explanation:

just did it on a.pex

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All boxes with a square​ base, an open​ top, and a volume of 60 ft cubed have a surface area given by ​S(x)equalsx squared plus
Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval (0,\infty) is S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

The dimensions of the box with minimum surface​ area are: the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

Step-by-step explanation:

We are given the surface area of a box S(x)=x^2+\frac{240}{x} where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval (0,\infty) and the dimensions of the box with minimum surface​ area.

1. To find the absolute minimum you must find the derivative of the surface area (S'(x)) and find the critical points of the derivative (S'(x)=0).

\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}

Next,

2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120

There is a undefined solution x=0 and a real solution x=2\sqrt[3]{15}. These point divide the number line into two intervals (0,2\sqrt[3]{15}) and (2\sqrt[3]{15}, \infty)

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before x=2\sqrt[3]{15}, increases after it, and is defined at x=2\sqrt[3]{15}. So f(x) has a relative minimum point at x=2\sqrt[3]{15}.

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for x=2\sqrt[3]{15}.

\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}

and for x=2\sqrt[3]{15} we get:

2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0

Therefore S(x) has a minimum at x=2\sqrt[3]{15} which is:

S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

2. To find the third dimension of the box with minimum surface​ area:

We know that the volume is 60 ft^3 and the volume of a box with a square base is V=x^2h, we solve for h

h=\frac{V}{x^2}

Substituting V = 60 ft^3 and x=2\sqrt[3]{15}

h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft

The dimension are the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

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A line has the equation y=x. It is translated up by 3 units. What is the equation of the new line?
olga nikolaevna [1]
Since we are shifting equation up 3 units, the equation would be y=x+3
4 0
3 years ago
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I have 2 questions 1.Matt needs to buy 13 cases of oil. Each case contains 24 bottles. How many bottles of oil will Matt have?
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Answer:

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Step-by-step explanation:

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Is v = 12 a solution to the inequality below?<br> V&gt; 7
KengaRu [80]

Answer: Yes

Step-by-step explanation: If V=12

Then we plug in for the inequality and get 12>7

7 0
3 years ago
Solve the following using pemdas 3 x 9 + 7
MAXImum [283]
27+7 = 34

so, the answer is 34
8 0
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