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Sliva [168]
3 years ago
6

Find the amount of interest owed for a $1,895 loan for 4 years at a 7.9% interest rate.

Mathematics
1 answer:
fiasKO [112]3 years ago
4 0

Answer: The amount of interest owed is $598.82.

Step-by-step explanation: 7.9% of $1,895 is 149.705. So, 149.705 x 4 years would be $598.82 for 4 years.

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Confidence Interval Mistakes and Misunderstandings—Suppose that 500 randomly selected recent graduates of a university were as
kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

n=500\\\bar x=8.6\\\sigma=2.2

(a)

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)

(b)

The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

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

In this case the sample size is,

<em>n </em>= 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

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Mamont248 [21]
You have to multiply by the complex conjugate so that you can clear the radical from the denominator. Once you do the multiplication simplify the radical. There's still a common factor in -54, -18, & 75 so divide all those by 3.

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Please Help Asap!!! I hate integers!!!
sveta [45]
5 times (-17)= -85 so you should have said -85
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