Simplify the following expression. X^2/9 . X^-4/9 x^2/9 . X^-4/9

Vina has $4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than qua

DanielleElmas [232]

Nickels (0.05): Q + 6

Dimes (0.10): Q + 8

Quarters (0.25): Q

Nickels + Dimes + Quarters = Total

0.05(Q + 6) + 0.10(Q + 8) + 0.25(Q) = 4.70

0.05Q + 0.30 + 0.10Q + 0.80 + 0.25Q = 4.70

0.40Q + 1.10 = 4.70

0.40Q = 3.60

Q = 9

Nickels (0.05): Q + 6 = (9) + 6 = 15

Dimes (0.10): Q + 8 = (9) + 8 = 17

Answer: Nickels = 15, Dimes = 17, Quarters = 9

4 0

3 years ago

01m/= 4x18.m/U = 30, m/s = 34. HT = 13. UN = Sy - 12. OT = 23.

Paul [167]

Answer:

its 23

Step-by-step explanation:

7 0

3 years ago

We would like to create a confidence interval.

Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

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 $1-\alpha$ , we have the confidence interval with a margin of error of:

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of $\sqrt{\pi(1-\pi)}$

So

a.A 99% confidence level and a sample size of 50 subjects.

$n = 50$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}$

b.A 90% confidence level and a sample size of 50 subjects.

$n = 50$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}$

c.A 90% confidence level and a sample size of 300 subjects.

$n = 300$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}$

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

$n = 300$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}$

6 0

3 years ago

An acid is prepared by mixing 400 mL of water and 100 mL of pure sulfuric acid. What is the

Natasha2012 [34]

C. 25
——-
Divide 400/100=4
——-
Than divide 4/100
—-

8 0

3 years ago

A newborn child receives a $20,000 gift toward college education from her grandparents. How much will the 20,000 be worth in 17

Alexus [3.1K]

Answer:

The amount the $20.000 will be worth in 17 years at compound interest is $65068.443

Step-by-step explanation:

Here we have the Principal, P = $20,000.00

The annual interest rate, r = 7% = 0.07

Time , t = 17 years

Number of compounding period per year, m = quarterly = 4

The compound interest can be found from the following formula;

$Amount, \ A = P \left (1 + \frac{1}{r} \right )^{mt}$

Therefore, by plugging the values of the equation parameters, we have;

$Amount, \ A = 20000 \left (1 + \frac{0.07}{4} \right )^{4 \times 17} = \$ 65068.443$

Therefore, the amount the $20.000 will be worth in 17 years at compound interest = $65068.443.

7 0

3 years ago