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otez555 [7]
3 years ago
13

A friend is building a dog pen with an area of 150 square feet. Each side must be at least 5 feet long.

Mathematics
1 answer:
JulijaS [17]3 years ago
6 0
The dog pen is most likely a rectangle. Area of a rectangle = length * width
a) possible dimension
150 / 5 = 30 ; length = 30 ; width = 5
150 / 7.5 = 20 ; length = 20 ; width = 7.5
150 / 10 = 15 ; length = 15 ; width = 10
150/ 12.5 = 12 ; length = 12.5 ; width = 10

b) maximum amount of fence. Compute for the perimeter.
P = 2(30 + 5) = 2(35) = 70
P = 2(20 + 7.5) = 2(27.5) = 55
P = 2(15 + 10) = 2(35) = 70
P = 2(12.5 + 12) = 2(24.5) = 49

MAXIMUM AMOUNT OF FENCE REQUIRED IS 70 FEET.

c) either: 30 by 5 or 15 by 10 dimension.

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6 0
4 years ago
If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w
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Let x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11} be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.

2. If the mean weight of the 7 other players is 202 lb, then

\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.

3. From the previous statements you have that

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Add these two equalities and then divide by 11:

x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.

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