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2 years ago

Find the range of f

Mathematics

1 answer:

Ugo [173]2 years ago

8 0

Write down y=f(x) and then solve the equation for x, giving something of the form x=g(y).
Find the domain of g(y), and this will be the range of f(x). ...
If you can't seem to solve for x, then try graphing the function to find the range.

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Marco picked 8

Rzqust [24]

It 4 the answer is 4 alpeass

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3 years ago

Mike and Beth both write the decimal 0.8 (0.88888...)as a fraction. Mike: 45 Beth: 89 Which student wrote the correct fraction?

fenix001 [56]

Answer:

see the explanation

Step-by-step explanation:

we have

0.888...

This is a <u>repeating decimal</u> (Is a decimal that has a digit, or a block of digits, that repeat over and over and over again without ever ending)

Convert to fraction number

Let

x=0.888...

10x=8.888...

Subtract 0.888... from 8.888... to remove the decimal

10x-x=8.888...-0.888...

9x=8

Solve for x

x=8/9

therefore

Mike fraction is incorrect

because 4/5=0.8

0.8 is a <u>terminating decimal </u>(It's a decimal with a finite number of digits)

Mike's mistake was considering the number as a terminating decimal instead of a repeating decimal

Beth is correct

because

If you divide 8/9

the result is 0.8888888...

3 0

2 years ago

Enter the sum of numbers as a product of their GCF.

frutty [35]

Their GCF is 6 and the sum is 66

Step-by-step explanation:

8 0

2 years ago

Look at the question below

ss7ja [257]

Answer:

y=6x is the answer

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

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:

$df=n-1=36-1=35$

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(0.005,35)".And we see that $t_{\alpha/2}=2.72$

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

$62-2.72\frac{8}{\sqrt{36}}=58.373$

$62+2.72\frac{8}{\sqrt{36}}=65.627$

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0

3 years ago