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xenn [34]
3 years ago
11

The weights, in pounds, of the upper-class men offensive line players for a college are shown below: 322, 290, 321, 326, 330, 31

5, 311, 292, 290 What is the average, or mean, weight of the upper-class men offensive players? Provide your answer with 3 sig figs. -311 pounds -315 pounds -290 pounds -318 pounds
Mathematics
2 answers:
I am Lyosha [343]3 years ago
8 0

Answer: 311 pounds

Step-by-step explanation:

To find the average, add up all of the values and divide by the amount of values.

First add up all of the values.

322+290+321+326+330+315+311+292+290 = 2797

Now count how many values there are and divide your previous sum by it.

2797/9 = 310.78

Because you need to have 3 sig figs, round to the nearest one.

310.78 rounds up to 311.

amm18123 years ago
8 0

answer-311

Step-by-step explanation:

322+290+321+326+330+315+311+292+290=2797

Count the number of values and divide by 2797

2797 divided by 9 =310.77777777777777

Nearest one-311

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You are designing a rectangular garden. the garden will be enclosed by fencing on three sides and by a house on the fourth side.
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Thus, the length of fencing needed is given by

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The area of the rectangle is given by xy,

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xy = 288  \\  \\  \\  \\ \Rightarrow y= \frac{288}{x}

Substituting for y into the equation for the length of fencing needed, we have

P=x+2\left( \frac{288}{x} \right)=x+ \frac{576}{x}

For the amount of fencing to be minimum, then

\frac{dP}{dx} =0 \\  \\ \Rightarrow1- \frac{576}{x^2} =0 \\  \\ \Rightarrow \frac{576}{x^2} =1 \\  \\ \Rightarrow x^2=576 \\  \\ \Rightarrow x=\sqrt{576}=24

Now, recall that

y= \frac{288}{x} = \frac{288}{24} =12

Thus, the length of fencing needed is given by

P = x + 2y = 24 + 2(12) = 24 + 24 = 48.

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3 years ago
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