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4 years ago

1. Find f(x) if it is known that f(x+5)=x+12.

Mathematics

1 answer:

uranmaximum [27]4 years ago

7 0

Answer:

1. f(x)=x+7

2. f(x)=1.2x

Step-by-step explanation:

1. If f(x+5)=x+12, then substitute

$t=x+5\Rightarrow x=t-5$

Hence,

$f(t)=(t-5)+12\\ \\f(t)=t+7$

Now change t into x:

$f(x)=x+7$

2. If f(x-5)=1.2x-6, then substitute

$t=x-5\Rightarrow x=t+5$

Hence,

$f(t)=1.2(t+5)-6\\ \\f(t)=1.2t+6-6\\ \\f(t)=1.2t$

Now change t into x:

$f(x)=1.2x$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers

The shape of the distribution of the time required to get an oil change at a 20 minute oil change facility. However, records ind

Katyanochek1 [597]

Answer:

For a mean oil change time of 20.51 minutes there would be a 10% chance of being at or below

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 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.

In this problem, we have that:

$\mu = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166$

Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below?

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

By the Central Limit Theorem

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

$-1.28 = \frac{X - 21.3}{0.6166}$

$X - 21.3 = -1.28*0.6166$

$X = 20.51$

For a mean oil change time of 20.51 minutes there would be a 10% chance of being at or below

6 0

4 years ago