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Dmitriy789 [7]
3 years ago
13

Leo bought a used car for x dollars. One year later the value of the car was 0.88x. Which expression is another way to describe

the change in the value of the car?

Mathematics
1 answer:
nalin [4]3 years ago
4 0
It would be a 12% decrease. 0.88 is 88% (multiply by 100 to find the percentage) of the original price (x, which is 100%). as an example, say x=1000. 

0.88(1000)=880

880 is 88% of 1000, but the original price did not decrease by 88%/$880. Instead subtract this percentage from 100. 

100%-88%=12% 

there, you find the percentage lost/the percentage of the value decrease.
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A racing car consumes a mean of 100 gallons of gas per race with a variance of 64. If 44 racing cars are randomly selected, what
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83.89% probability that the sample mean would be greater than 98.8 gallons

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

The standard deviation is the square root of the variance. So

\mu = 100, \sigma = \sqrt{64} = 8, n = 44, s = \frac{8}{\sqrt{44}} = 1.21

If 44 racing cars are randomly selected, what is the probability that the sample mean would be greater than 98.8 gallons

This probability is 1 subtracted by the pvalue of Z when X = 98.8. So

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

By the Central Limit Theorem

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

Z = \frac{98.8 - 100}{1.21}

Z = -0.99

Z = -0.99 has a pvalue of 0.1611

1 - 0.1611 = 0.8389

83.89% probability that the sample mean would be greater than 98.8 gallons

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3 years ago
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