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Nikitich
2 days ago
7

A company that makes fleece clothing uses fleece produced from two farms, Northern Farm and Western Farm. Let the random variabl

e X represent the weight of fleece produced by a sheep from Northern Farm. The distribution of X has a mean 14.1 pounds and a standard deviation 1.3 pounds. Let the random variable Y represent the weight of fleece produced by a sheep from Western Farm. The distribution of Y has a mean 6.7 pounds and a standard deviation of 0.5 pounds. Assume X and Y are independent. Let W equal the total weight of fleece from 10 randomly selected sheep from Northern Farm and 15 randomly selected sheep from Western Farm. Which of the following is the standard deviation, in pounds, of W?
A) 1.3+0.5
B) sqrt(1.3^2+0.5^2)
C) sqrt(10(1.3)^2+15(0.5)^2)
D) sqrt(10^2(1.3)^2+15^2(0.5)^2)
E) sqrt((1.3)^2/10 + (0.5)^2/15
Mathematics
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The effective annual interest rate calculations yield:
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8 0
2 months ago
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At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A
zzz [12365]

Response:

Part A

The best choice is;

b. y = 128(0.989)x + 72

Part B

Ella can consume the tea post 7:22

Detailed clarification:

Here, we observe the temperature variation with time represented in an exponential equation as follows;

Let the temperature at time x (measured in minutes) be y

Thus, y = a × mˣ + b

Where:

b = Curve shift or the limiting value of the decreasing exponential function as x → ∞

When y = 200, x = 0

This leads to 200 = a × m⁰ + c = a + c

Here, c represents the graph's shift, which is the temperature increase = final temperature = 72°F

Thus, a = 200 - 72 = 128°F

When y = 197, at x = 2 minutes

Thus, 197 = 128·m² + 72 =

m² = (197 - 72)/128 = 125/128

m = √(125/128) = 0.98821

Therefore, the exponential cooling equation is given by;

y = 128 × (0.98821)ˣ + 72

Hence the best choice is b. y = 128(0.989)x + 72

Part B

When the tea's temperature reaches 172°F, we have;

172 = 128 × (0.989)ˣ + 72

Thus, (0.989)ˣ = (172 - 72)/128 = 100/128 = 25/32

log(0.989)ˣ = log(25/32)

x·log(0.989) = log(25/32)

x = log(25/32)/log(0.989) = 22.32 minutes

Thus, Ella can consume the tea post 7:22.

6 0
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36 divided by 6
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