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2 years ago

8

The value of a boat is $23,400. It loses 8% of its value every year. Find the approximate monthly percent decrease in value. Rou

nd your answer to the nearest hundredth of a percent.

1 answer:

Oduvanchick [21]2 years ago

8 0

23400 x 8/100 = 1872 = the loss
1872 : 12 = 156= the loss each month
156/1872*100% = 8,33 % then round it

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Consider a sampling distribution with p equals 0.15p=0.15 and samples of size n each. Using the appropriate formulas, find the

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Answer:

$a.\ \mu_p=750\ \ , \sigma_p=0.005\\\\b.\ \mu_p=150\ \ , \sigma_p=0.0113\\\\c.\ \mu_p=75\ \ , \sigma_p=0.0160$

Step-by-step explanation:

a. Given p=0.15.

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$\mu_p=np\\\\=0.15\times 5000\\\\=750$

-The standard deviation is calculated using the formula:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\=\sqrt{\frac{0.15(1-0.15)}{5000}}\\\\=0.0050$

Hence, the sample mean is μ=750 and standard deviation is σ=0.0050

b. Given that p=0.15 and n=1000

#The mean of a sampling proportion of n=1000 is calculated as:

$\mu_p=np\\\\=1000\times 0.15\\\\\\=150$

#-The standard deviation is calculated as follows:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{1000}}\\\\\\=0.0113$

Hence, the sample mean is μ=150 and standard deviation is σ=0.0113

c. For p=0.15 and n=500

#The mean is calculated as follows:

$\mu_p=np\\\\\\=0.15\times 500\\\\=75$

#The standard deviation of the sample proportion is calculated as:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{500}}\\\\\\=0.0160$

Hence, the sample mean is μ=75 and standard deviation is σ=0.0160

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Step-by-step explanation:

10/25= x/100

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balu736 [363]

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I believe that it is b

Step-by-step explanation:

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marshall27 [118]

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