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lisabon 2012 [21]
2 years ago
3

A bicycle depreciates at a rate of 15% therese bought a bicycle for $250. How much should it be worth 6 years later ?

Mathematics
1 answer:
Virty [35]2 years ago
3 0

Answer:

\$94.29  

Step-by-step explanation:

we know that

The  formula to calculate the depreciated value  is equal to  

V=P(1-r)^{x}  

where  

V is the depreciated value  

P is the original value  

r is the rate of depreciation  in decimal  

x  is Number of Time Periods  

in this problem we have  

P=\$250\\r=15\%=0.15\\x=6\ years

V=\$250(1-0.15)^{6}  

V=\$250(0.85)^{6}=\$94.29  

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Create a set of data that contains 11 test scores and satisfies each condition below:Mean: 83Median: 81Mode: 80Range: 26
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2 years ago
The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th
Feliz [49]

Answer:

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, we need 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, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

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

By the Central Limit Theorem

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

Z = \frac{30.9 - 29}{0.8341}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

X = 27.1

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

Z = \frac{27.1 - 29}{0.8341}

Z = -2.28

Z = -2.28 has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

7 0
2 years ago
The following table shows the salary of 7 people in an office last year.
kotegsom [21]

Answer:

a.  £24,714.29

b. £16,833.33

Step-by-step explanation:

The calculation of mean income is given below:-

Mean income = Total addition of salaries ÷ Number of workers

= £9,500 + £25,000 + £13,250 + £72,000 + £12,750 + £29,500 + £11,000

= £173,000 ÷ 7

= £24,714.29

Now,

the Mean income excluding Deva's salary:

= Formula of Mean income

= Total addition of salaries excluding Deva salary ÷ Number of workers

= (£9,500 + £25,000 + £13,250 + £12,750 + £29,500 + £11,000) ÷ 6

= £101,000 ÷ 6

= £16,833.33

7 0
2 years ago
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