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3 years ago

For the function f(x) = x + 9, find (f o f-1)(5) A. 14 B. 5 C. -5 D. 25

1 answer:

5 0

F-1:
y=x+9
y-9=x
x-9=f-1
f-1(14)=14-9=5 f-1(5)=5-9=-4 f-1(-5)=-5-9=-14 f-1(25)=25-9=16
f(14)=14+9=24 f(5)=5+9=14 f(-5)=9-5=4 f(25)=25+9=34

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The 99% confidence interval would be given by (196.2;283.8)

We are 99% condident that the true mean is between 196.2 and 283.8

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=240$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=15 represent the sample standard deviation

n=4 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=4-1=3$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,3)".And we see that $t_{\alpha/2}=5.84$

Now we have everything in order to replace into formula (1):

$240-5.84\frac{15}{\sqrt{4}}=196.2$

$240+5.84\frac{15}{\sqrt{4}}=283.8$

So on this case the 99% confidence interval would be given by (196.2;283.8)

We are 99% condident that the true mean is between 196.2 and 283.8

What assumption about the data is necessary for the inference derived from the analysis to be valid?

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.

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