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3 years ago

Let f(x)=7x^2-8. find f(a^2)

Mathematics

1 answer:

strojnjashka [21]3 years ago

6 0

Answer:

$\large\boxed{f(a^2)=7a^4-8}$

Step-by-step explanation:

$f(x)=7x^2-8\\\\f(a^2)\to \text{put}\ x=a^2\ \text{to the equation of the function}\ f:\\\\f(a^2)=7(a^2)^2-8\qquad\text{use}\ (a^n)^m=a^{nm}\\\\f(a^2)=7a^{2\cdot2}-8\\\\f(a^2)=7a^4-8$

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Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbar

Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : $n=45$

$\overline{x}=148.79\text{ lb}$

$\sigma=31.37\text{ lb}$

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = $\mu=148.79\text{ lb}$

b) The confidence interval for the population mean is given by :-

$\mu\ \pm E$ , where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Margin of error : $E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69$

Now, the 90% confidence interval for the population mean will be :-

$148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)$

Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0

3 years ago

1.) A line has a slope of 2 and passes through the point (3,4). Write the equation of the line in slope-intercept form. [y=mx+b]

emmasim [6.3K]

Answer: y= 2x-2

Step-by-step explanation:

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3 years ago

HELP ASAP(FIRST ANSWER GETS BRAINLIST)!

quester [9]

Answer:

I would say B, but can't be sure because is part of the question cut off?

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Combine the terms in the following expressions. a. 10x-4x

Ilya [14]

Answer:

Step-by-step explanation:

Step 1:

10x - 4x

Answer:

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3 years ago

An aptitude test is designed to measure leadership abilities of the test subjects. Suppose that the scores on the test are norma

Komok [63]

Using the normal distribution, it is found that 0.0329 = 3.29% of the population are considered to be potential leaders.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

In this problem:

The mean is of 550, hence $\mu = 550$ .

The standard deviation is of 125, hence $\sigma = 125$ .

The proportion of the population considered to be potential leaders is 1 subtracted by the p-value of Z when X = 780, hence:

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

$Z = \frac{780 - 550}{125}$

$Z = 1.84$

$Z = 1.84$ has a p-value of 0.9671.

1 - 0.9671 = 0.0329

0.0329 = 3.29% of the population are considered to be potential leaders.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213

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2 years ago