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3 years ago

A boat costs $11850 and decreases in value by 10% per year how much will the boat be worth after 8 years

Mathematics

2 answers:

Galina-37 [17]3 years ago

8 0

Answer: the boatf will be worth $5101 after 8 years.

A decrease of 10% is equivalent to multiplying the current value by 0.9.

Every year the value gets multiplied so we get an exponential function in years:

value after y years = $V\cdot0.9\cdot0.9\cdot\cdot\cdot0.9=V\cdot0.9^y$

value after 8 years = $\$11850\cdot0.9^8=\$11850\cdot0.43\approx \$5101$

ziro4ka [17]3 years ago

4 0

Answer:

Step-by-step explanation:

Keep subtracting 10 %

U will get 5101 approx.

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If the function 5x+y=1 has the domain (-2,1,6) then what is the corresponding range?

zysi [14]

The corresponding range of the domain is (11, -4, -29)

<h3>How to determine the range?</h3>

The function is given as:

5x + y = 1

Make y the subject

y = 1 - 5x

The domain is given as:

(-2,1,6)

Substitute these values in y = 1 - 5x

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y = 1 - 5(1) = -4

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Hence, the corresponding range of the domain is (11, -4, -29)

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1 year ago

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2 years ago

A manufacturer produces crankshafts for an automobile engine. the crankshafts wear after 100,000 miles (0.0001 inch) is of inter

Daniel [21]

Part A:

Significant level:

α = 0.05

Null and alternative hypothesis:

h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{2.78-3}{0.9/\sqrt{15}} \\ \\ = \frac{-0.22}{0.2324} =-0.9467$

P-value:

P(-0.9467) = 0.1719

Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438

Conclusion:

Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.

Part B:

The power of the test is given by:

$\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\ \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\ \\ =0.8117-0.0012=0.8105$

Therefore, the power of the test if μ = 3.25 is 0.8105.

Part C:

The sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:

$1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\ \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\ \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\ \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898$

$\Rightarrow n=(3.8898)^2=15.13$

Therefore, the sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.

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3 years ago