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3 years ago

How do I simplify this problem?

Mathematics

1 answer:

Norma-Jean [14]3 years ago

6 0

I think you would just have to find the square roots

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Talib is trying to find the inverse of the function to the right. His work appears beneath it. Is his work correct? Explain your

aev [14]

The answer is is 3 hope it helps

6 0

3 years ago

Hey therapy was conducted to see how many people ride a bus to school every day. 160 people reported that they ride the bus to s

xeze [42]

Answer:

200 people

Step-by-step explanation:

Let: the number of people who surveyed be: x

Therefore, (160/x) x 100=80%

=> x=(160x100)/80

=>x=16000/80

=>x= 200

Thus, 200 people surveyed.

Mark as brainliest

4 0

3 years ago

If a number is an element of the set of real

kvasek [131]

Answer:

Option (A)

Step-by-step explanation:

A set of real numbers is a combination of both rational and irrational numbers.

But the elements in the sets of rational and irrational numbers will be different.

In short, rational and irrational number sets are the two subsets of real numbers.

Hence, the number is an element of either the set of rational numbers or the set of irrational numbers.

Option (A) is the correct option.

3 0

3 years ago

Jan spent the day at the mall. She came home with $9.50. She spent $5.25 on earrings and 3 times that amount on a shirt. How muc

ki77a [65]

Answer:

She took $30.50 to the mall originally

Step-by-step explanation:

Let us solve the problem step by step

∵ She spent $5.25 on earrings

∵ The price of the shirt is 3 times the price of the earrings

∴ The price of the shirt = 3 × 5.25 = 15.75

∴ She spent $15.75 on the shirt

Find her total spending in the mall by adding the prices of earrings and the shirt

∴ She spent = 5.25 + 15.75 = 21

∴ She spent $21 on the mall

∵ She came home with $9.50

∵ She spent $21 on the mall

- Add them to find her original amount of money

∴ She took = 21 + 9.50 = 30.50

∴ She took $30.50 to the mall originally

5 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago