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3 years ago

a boat cost 92,000.00 and depreciates in value by 15% per year how much will the boat be worth after 10 years

Mathematics

2 answers:

Reil [10]3 years ago

8 0

Guessing you are talking about compounding value here. In that case, it should be worth around 18,100 dollars.

UNO [17]3 years ago

8 0

A=p (1-r)^t
A=92,000×(1−0.15)^(10)
A=18,112

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From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

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Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

$\= x = 13.4$ . $E = 2.3$

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

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3 years ago