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AfilCa [17]
2 years ago
12

Dylan purchased a new car in 1993 for $26,300. The value of the car has been depreciating exponentially at a constant rate. If t

he value of the car was $2,700 in the year 2001, then what would be the predicted value of the car in the year 2006, to the nearest dollar?
Mathematics
1 answer:
emmainna [20.7K]2 years ago
8 0

The predicted value of the car in the year 2006 to the nearest dollar would be  $651.

<h3>What is the predicted value of the car?</h3>

The first step is to determine the rate of depreciation

g = (FV/PV)^(1/n) - 1

Where:

FV = value of the car in 2001

PV = value of the car in 1993

n = number ofyears = 8

(2700/26,300)^(1/8) - 1  = -24.76%

Now determine the value of the car in 2006

2700x ( 1 - 0.2476)^5 = $651

To learn more about depreciation, please check: brainly.com/question/25552427

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Answer:

Step-by-step explanation:

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Svetach [21]

Using the normal distribution, it is found that 25.14% of the batteries will last more than 420 hours.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
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In this problem, we have that the mean and the standard deviation are given, respectively, by:

\mu = 400, \sigma = 30.

The proportion of the batteries will last more than 420 hours is <u>one subtracted by the p-value of Z when X = 420</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{420 - 400}{30}

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0.2514 = 25.14% of the batteries will last more than 420 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

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2 years ago
Evaluate the expression for the given value of variable(s)
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Answer:

Total area:75

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