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3 years ago

5

The next model of a sports car will cost 13.6% more than the current model. The current model costs $58,000 . How much will the

price increase in dollars? What will be the price of the next model?
Increase price:

Price of next model:

1 answer:

Anettt [7]3 years ago

7 0

Answer:

increase price: $7,888

price of next model:65,888

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2 years ago

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3 years ago

Read 2 more answers

A random sample of 28 statistics tutorials was selected from the past 5years and the percentage of students absent from each one

Vesna [10]

Using the t-distribution, the 90% confidence interval for the mean number of absences is given by: (9.22, 11.60).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

t is the critical value.

n is the sample size.

s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 28 - 1 = 27 df, is t = 1.7033.

Researching this problem on the internet, the parameters are given by:

$\overline{x} = 10.41, s = 3.71, n = 28$ .

Hence the bounds of the interval are given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 10.41 - 1.7033\frac{3.71}{\sqrt{28}} = 9.22$

$\overline{x} + t\frac{s}{\sqrt{n}} = 10.41 + 1.7033\frac{3.71}{\sqrt{28}} = 11.60$

The 90% confidence interval for the mean number of absences is given by: (9.22, 11.60).

More can be learned about the t-distribution at brainly.com/question/16162795

#SPJ1

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2 years ago

The number of winter storms in a good year is a Poisson random variable with mean 3, whereas the number in a bad year is a Poiss

djyliett [7]

Answer: Mean = 4.8 and variance = 5.16

Step-by-step explanation:

Since we have given

Let X be the number of storms occur in next year

Y= 1 if the next year is good.

Y=2 if the next year is bad.

Mean for good year = 3

probability for good year = 0.4

Mean for bad year = 5

probability for bad year = 0.6

So, Expected value would be

$E[x]=\sum xp(x)\\\\=3\times 0.4+5\times 0.6\\\\=1.2+3\\=4.2$

Variance of the number of storms that will occur.

$Var[x]=E[x^2]-(E[x])^2$

$E[x^2]=E[x^2|Y=1].P(Y=1)+E[x^2|Y=2].P(Y=2)\\\\=(3+9)\times 0.4+(5+25)\times 0.6\\\\=12\times 0.4+30\times 0.6\\\\=4.8+18\\\\=22.8$

So, Variance would be

$\sigma^2=22.8-(4.2)^2\\\\=5.16$

Hence, Mean = 4.8 and variance = 5.16

7 0

3 years ago