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Phantasy [73]
3 years ago
9

The initial value of a car is $27,000. After one year, the value of the car is $20,250. What exponential function models the exp

ected value of the car? Estimate the value of the car after 3 years.
A. $13,824
B. $11,391
C. $16,581
D. $19,683
Mathematics
1 answer:
nasty-shy [4]3 years ago
3 0
P=20250/27000=0.75,,,y=a*b^t Y=27000(0.75)^3=11,390.625
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If 70% of college students say they have a job what is the probability that 3 randomly selected college students say they have a
dsp73

Answer: 0.343

Step-by-step explanation:

Probability of college students that say they have a job = 70%

Number of randomly selected college students = 3

The probability that 3 randomly selected college students say they have a job will be:

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4 0
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What is the equation for the line that passes through coordinates: (2,7
Agata [3.3K]

Answer:

Y=3x+1

Step-by-step explanation:

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3 years ago
What is the closest perfect square greater than 54
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Answer:

64

Step-by-step explanation:

Perfect squares are integers multiplied by themselves.

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  • 3 times 3  = 9
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The closest perfect squares to 54 are 49 (7^2) and 64 (8^2).

49 is less than 54, so that's ruled out.

Therefore, the closest perfect square to 54 that is greater than it is 64.

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How many values are between 30 and 50?
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2 years ago
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc
Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

M = z*\frac{\sigma}{\sqrt{n}}

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

n = 718.57

Rouding up

We need a sample size of at least 719

6 0
3 years ago
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