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Hatshy [7]
3 years ago
5

in an over-fished area, the catch of a certain type of fish is decreasing at an average rate of 8% per year. If this decline per

sists, how long will it take for the catch to reach half of the amount before the decline???
Mathematics
1 answer:
Galina-37 [17]3 years ago
4 0
<span>n = n0(1 - 0.08)^t 
= n0(0.92)^t 

Putting n = n0 / 2: 
1 / 2 = 0.92^t 
t = log(1 / 2) / log(0.92) 
= 8.31 yr.</span><span>
</span>
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The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

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And the standard deviation of the distribution of the sum of values of X is given by,  

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The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

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\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

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