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I h(x)is the inverse of f (x)what is the value of h (f (x))

sergeinik [125]

Answer:

h(f(x)) = x

Step-by-step explanation:

Since the functions are inverses h(x) undoes f(x)

That is if f(x) = x then inverse function maps back to x

h(f(x)) = x

6 0

2 years ago

Write two addition equations that each have a sum of -21.5

blondinia [14]

Answer:

-8.5 + -13 = -21.5

and

-15.5 + -6 = -21.5

6 0

2 years ago

A spinner has 4 equally-sized sections, labeled A, B, C, and D. It is spun and a fair coin is tossed. What is the denominator of

jekas [21]

32% i believe because i added 4 an 2 together from the spinner an coin an got 6 the i divided 1 by 6 to get .16 and then doubled it because the precents before that doesnt matter if your looking for a specific combination

4 0

3 years ago

Read 2 more answers

A student received a coupon for 13% off the total purchase price at a clothing store. Let C be the original price of the purchas

frez [133]

Answer:

C-0.13C=0.87C

Step-by-step explanation:

the whole - 13%=87%

3 0

2 years ago

What sample size is needed to give a margin of error within in estimating a population mean with 95% confidence, assuming a prev

atroni [7]

Answer:

$n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion $\pi$ (if no previous estimate use 0.5) and M is the desired margin of error.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Needed sample size:

The needed sample size is n. We have that:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{\pi(1-\pi)}{n}}$

$\sqrt{n}M = 1.96\sqrt{\pi(1-\pi)}$

$\sqrt{n} = \frac{1.96\sqrt{\pi(1-\pi)}}{M}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

$n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion $\pi$ (if no previous estimate use 0.5) and M is the desired margin of error.

4 0

2 years ago