Simplify the given expression. Assume x doesn’t = 0. (20x^-3/10x^-1)^-2

Perry made $20,000 per year.He soon he received a promotion(raise) with a 20% increase in pay.What is his new salary?How much is

user100 [1]

Hello! The salary goes up 20% due to the promotion. His original salary was $20,000. To find out his new salary, you have to do is multiply the original price by the rate. 20,000 * 20% (0.2) is 4,000. His promotion is $4,000. Now, we add. 20,000 + 4,000 is 24,000. There. Perry's new salary is $24,000.

5 0

3 years ago

20 POINTS PLEASE HELP

Andru [333]

The figure RQTS is a clockwise rotation of figure DEFC, then:
The image of CD is
Answer: Option C. RS

8 0

3 years ago

Can somebody help me??

Dominik [7]

Answer:

1. -72

2. -64

Step-by-step explanation:

it is increasing by 4 so,

-76, -72, -68, -64, -60, -56.

4 0

3 years ago

3х – 8y = — 16 = M So yeah I kinda need help

marishachu [46]

Answer:

3x-8y-16 = 0

Step-by-step explanation:

8 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago