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cestrela7 [59]
3 years ago
13

In 2011, the moose population in a park was measured to be 5,700. By 2018, the population was measured

Mathematics
1 answer:
Naily [24]3 years ago
5 0

Answer:

P = 5700(1.074ⁿ)

13,437 moose in 2023    13,425 if you use the rounded function above.

Step-by-step explanation:

p = 5700(1 + x)ⁿ

n = years elapsed     x = percentage increase per year

2018 - 2011 = 7 years

9400 = 5700(1 + x)⁷

9400/5700 = (1 + x)⁷

ln(9400/5700) = 7ln(1 + x)

ln(9400/5700) / 7 = ln(1 + x)

e^(ln(9400/5700) / 7) = e^(ln(1 + x))

1.07407879 = 1 + x

x = 0.07407879

2023 - 2011 = 12

P = 5700(1.074¹²) = 13,437

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The supreme shipping company can load trucks with both rectangular and cylindrical containers. A rectangular container has a vol
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the number of rectangular containers to maximze their income is 18 while the number of cylindrical containers to maximize income is 12.

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Let x be the number of rectangular containers and y the number of cylindrical containers.

Since a rectangular container has a volume of 100 ft ³and weighs 200 pounds and a cylindrical container has a volume of 200 ft.³ and weighs 100 pounds, and each truck has room for at most 4200 ft.³ of containers and can carry a maximum of 4800 pounds,

We have that the maximum volume of the truck V = 100x + 200y.

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Since V = 4200 ft.³  and W = 4800 pounds,

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So, the number of rectangular containers to maximze their income is 18 while the number of cylindrical containers to maximize income is 12.

The maximum income is $1740

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