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Nataliya [291]
3 years ago
14

A major fishing company does its fishing in a local lake. The first year of the company's operations it managed to catch 115,000

fish. Due to population decreases, the number of fish the company was able to catch decreased by 6% each year. How many total fish did the company catch over the first 20 years, to the nearest whole number? ​
Mathematics
1 answer:
Afina-wow [57]3 years ago
3 0

Answer:

Rounded to the nearest whole number, the company caught 1,360,630 fish in total over the 20 years.

Step-by-step explanation:

Given that a major fishing company does its fishing in a local lake, and the first year of the company's operations it managed to catch 115,000 fish but due to population decreases, the number of fish the company was able to catch decreased by 6% each year, to determine how many total fish did the company catch over the first 20 years, to the nearest whole number, the following calculation must be performed:  

Year 1: 115,000

Year 2: 115,000 x 0.94 = 108,100

Year 3: 108,100 x 0.94 = 101,614

Year 4: 101,614 x 0.94 = 95,517.16

Year 5: 95,517.16 x 0.94 = 89,786.13

Year 6: 89,786.13 x 0.94 = 84,398.96

Year 7: 84,398.96 x 0.94 = 79,335.02

Year 8: 79,335.02 x 0.94 = 74,574.92

Year 9: 74,574.92 x 0.94 = 70,100.42

Year 10: 70,100.42 x 0.94 = 65,894.40

Year 11: 65,894.40 x 0.94 = 61,940.73

Year 12: 61,940.73 x 0.94 = 58,824.29

Year 13: 58,824.29 x 0.94 = 54,730.83

Year 14: 54,730.83 x 0.94 = 51,446.98

Year 15: 51,446.98 x 0.94 = 48,360.16

Year 16: 48,360.16 x 0.94 = 45,458.55

Year 17: 45,458.55 x 0.94 = 42,731.03

Year 18: 42,731.03 x 0.94 = 40,167.17

Year 19: 40,167.17 x 0.94 = 37,757.14

Year 20: 37,757.14 x 0.94 = 35,491.71

TOTAL: 1,360,629.60

Thus, rounded to the nearest whole number, the company caught 1,360,630 fish in total over the 20 years.

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Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

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If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

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Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

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0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

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Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

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