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Assoli18 [71]
2 years ago
15

The population of gray wolves in Wisconsin increased from 570 in 2007 to 660 in 2014. Find the percent increase in the wolf popu

lation.
Mathematics
1 answer:
melamori03 [73]2 years ago
7 0

Answer:

15.78947% increase

Step-by-step explanation:

First: work out the difference (increase) between the two numbers you are comparing.

Increase = New Number - Original Number

Then:  divide the increase by the original number and multiply the answer by 100.

% increase = Increase ÷ Original Number × 100

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Answer:

a) 0.0167

b) 0

c) 5.948

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.16 ounces

Standard Deviation, σ = 0.08 ounces

We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) Standard deviation of 23 bags

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b) P( fill volume of 23 bags is below 5.95 ounces)

P(x < 5.95)

P( x < 5.96) = P( z < \displaystyle\frac{5.95 - 6.16}{0.0167}) = P(z < -12.57)

= 1 - P(z < 12.57)

Calculation the value from standard normal z table, we have,  

P(x < 5.95) = 1 - 1 = 0

c) P( fill volume of 23 bags is below 6 ounces)  = 0.001

P(x < 6)  = 0.001

P( x < 6) = P( z < \displaystyle\frac{6 - \mu}{0.0167})

Calculation the value from standard normal z table, we have,  

P( z \leq -3.09) = 0.001

\displaystyle\frac{6 - \mu}{0.0167} = -3.09\\\\\mu = 6 + (0.0167\times -3.09) = 5.948

If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.

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N is the same as 17, so it equals 17.
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