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WITCHER [35]
3 years ago
7

If f(x) =

"absmiddle" class="latex-formula">, what is the equation for f–1(x)?
Mathematics
2 answers:
Delvig [45]3 years ago
7 0

Answer:

f^{-1}(x) = x^{2}-6x+9 is the answer.

Step-by-step explanation:

If the function is f(x) = √x + 3

We will write the function as an equation

y = √x + 3

√x = y - 3

x = (y -3)²

x = y² -6y +9

Now we will convert the equation into a function

f^{-1}(x) = x^{2}-6x+9

Therefore the inverse function f(x) = x²-6x+9

mart [117]3 years ago
4 0

Answer:

f⁻¹(x) = (x-3)²

Step-by-step explanation:

We have given a function.

f(x) = √x + 3

We have to find inverse of above function.

f⁻¹(x) = ?

Putting y = f(x) in above equation, we have

y = √x + 3

We have to separate x from above equation.

Adding -3 to both sides of above equation, we have

y-3 = √x+3-3

y-3 = √x

Taking square to both sides of above equation, we have

(y-3)² = (√x)²

(y-3)² = x

Swapping  above equation , we have

x = (y-3)²

Putting x = f⁻¹(y) in above equation, we have

f⁻¹(y) = (y-3)²

Replacing y with x , we have

f⁻¹(x) = (x-3)² which is the answer.

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The mean number of words per minute (WPM) read by sixth graders is 97 with a standard deviation of 19 WPM. If 75 sixth graders a
andrey2020 [161]

Answer:

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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 = 97, \sigma = 19, n = 75, s = \frac{19}{\sqrt{75}} = 2.1939

What is the probability that the sample mean would be greater than 101.63 WPM?

This is 1 subtracted by the pvalue of Z when X = 101.63. So

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

By the Central Limit Theorem

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

Z = \frac{101.63 - 97}{2.1939}

Z = 2.11

Z = 2.11 has a pvalue of 0.9826.

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

5 0
3 years ago
I got so many points, so i wanna give some away!!
MrRissso [65]

Answer:

10 thanks by the way

Step-by-step explanation:

Have a great day

3 0
3 years ago
Read 2 more answers
What is the solution to this equation?
labwork [276]
The solution to this equation is C (6)

3x - 3 = 15

3x = 15 + 3
3x = 18
3/3 x = 18/3

x = 6
6 0
2 years ago
Read 2 more answers
What is the answer to this problem that I am having trouble with?
Gnesinka [82]

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

Read the values from the graph in the usual way: find the x-coordinate, then locate the point on the graph and find its y-coordinate.

You find the y-coordinate by following the horizontal line to the y-axis, where the numbers are.

4 0
3 years ago
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
3 years ago
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