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3 years ago

6

a car salesman purchased a car at an auction for $8,550. he will re-sell the car at a markup of $11,970. what is the percent inc

rease of the markup?

2 answers:

nlexa [21]3 years ago

7 0

Answer:

40%

Step-by-step explanation:

You subtract 11,970-8,550 = 3,420 to isolate the increase.

Divide 3,420/8,550 = .4 to get the percent increase of the markup.

The answer is .4 or 40% (they are equivalent)

You can check my work by multiplying 8,550 * .4 = 3,420

This answer is the percent of 8,550 that it increased (40% of 8,550) so make sure that is what the question is asking. If it is then it's your answer.

earnstyle [38]3 years ago

3 0

Answer:

40%

Step-by-step explanation:

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Truer or false? subtracting 7 from the parent linear function results in a reflection in the x-axis

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A reflection in the x-axis is denoted by - f(x), so the answer is false

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1 year ago

Find the volume of the cylinder.<br> 1607.7 in 3<br> 2845.7 in 3<br> 6430.7 in 3<br> 401.9 in 3

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Answer:

$\Huge\boxed{1607.7in^3}$

Step-by-step explanation:

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The volume of a cylinder can be found using this formula

$V=\pi r^2h$

where r = radius and h = height

The cylinder shown has a radius of 8 inches and a height of 8 inches

Knowing this information we plug it in into the formula

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3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = <em>mean age of student cars.</em>

$\mu_2$ = <em>mean age of faculty cars.</em>

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, <u><em>the test statistics</em></u> = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

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