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myrzilka [38]
3 years ago
10

A delivery company purchased a new delivery truck for $43,200. The depreciated value of the truck can be estimated by the number

of t years in service. If the the truck depreciates at a rate of $7,200 per year, which of the functions ƒ(t) could be used to estimate the truck's depreciated value after t years?
Mathematics
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

f(t)=43,200-7200t

Step-by-step explanation:                      

We have been given that a delivery company purchased a new delivery truck for $43,200. The depreciated value of the truck can be estimated by the number of t years in service. The truck depreciates at a rate of $7,200 per year.

This means that after t years truck will depreciate 7200t. To find the truck's depreciated value after t years we will subtract 7200t from the cost of new delivery truck (43,200).  

f(t)=43,200-7200t          

Therefore, the function f(t)=43,200-7200t could be used to estimate the truck's depreciated value after t years.

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A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The s
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Answer:

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question, we have that:

\mu = 270, \sigma = 90, n = 18, s = \frac{90}{\sqrt{18}} = 21.2

What is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days?

This is the pvalue of Z when X = 260. So

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

By the Central Limit Theorem

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

Z = \frac{260 - 270}{21.2}

Z = -0.47

Z = -0.47 has a pvalue of 0.3192.

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

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You roll two fair dice, one green and one red. (a) Are the outcomes on the dice independent? Yes No (b) Find P(1 on green die an
navik [9.2K]

Answer:

1) yes ; 2) 1/36 ; 3) 1/36 ; 4) 1/18

Step-by-step explanation:

Given the following :

Two fair dice : 1 green ; 1 red

A) Are the outcomes on the dice independent:

Yes, becomes the outcome of the green dice does not have any effect on the outcome of the red dice.

B) Find P(1 on green die and 5 on red die).

Probability = (number of required outcome) / (total possible outcomes)

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P(1 on green die and 5 on red die) :

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C) Find P(5 on green die and 1 on red die)

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P(1 on red) = 1/6

Find P(5 on green die and 1 on red die):

1/6 × 1/6 = 1/36

D) Find P((1 on green die and 5 on red die) or (5 on green die and 1 on red die))

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P((1 on green die and 5 on red die) or (5 on green die and 1 on red die)) =

P(5 on green die and 1 on red die) + P(1 on green die and 5 on red die)

= (1/36 + 1/36) = 2 /36 = 1/18

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