All categories

2 years ago

Which of the following is equal to [(x2y3)-2/(x6y3x)2]3 ?

Mathematics

2 answers:

Alona [7]2 years ago

4 0

Answer:

a=√x3y5 or a=−√x3y5

Step-by-step explanation:

inn [45]2 years ago

4 0

Answer:

A, B & D

Step-by-step explanation:

Just did it on Edge

You might be interested in

Carlos is looking at moving into a new apartment. He found one that costs $750 per month, but they require that the cost of rent

Over [174]

Answer: 2250

Step-by-step explanation:

the smallest amount of money Carlos needs is 2250 because 750 is exactly 1/3 of 2250.

7 0

2 years ago

What is the solution to the inequality?<br><br> −4x−8>−20

Sliva [168]

<span>−4x−8>−20
Add 8 to both sides
-4x>-12
Divide both sides by -4
Final Answer: x<3</span>

4 0

3 years ago

A random sample of 200 shoppers at a local grocery store found that 135 of the 200 sampled

VMariaS [17]

Using the z-distribution, as we are working with a proportion, it is found that samples of 937 should be taken.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

The margin of error is given by:

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

In this problem, we have that:

The estimate is of $\pi = 0.675$ .

The margin of error is of M = 0.03.

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

Then, we solve for n to find the minimum sample size.

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

$0.03 = 1.96\sqrt{\frac{0.675(0.325)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.675(0.325)}$

$\sqrt{n} = \frac{1.96\sqrt{0.675(0.325)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.675(0.325)}}{0.03}\right)^2$

$n = 936.4$

Rounding up, it is found that samples of 937 should be taken.

More can be learned about the z-distribution at brainly.com/question/25890103

4 0

2 years ago

Limit xtens to 0 x^2logx^2 what is the ans of interminate forms?

GarryVolchara [31]

Rewrite the limit as

$\displaystyle\lim_{x\to0}x^2\log x^2=\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}$

Then both numerator and denominator approach infinity (with different signs, but that's not important). Applying L'Hopital's rule, we get

$\displaystyle\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}=\lim_{x\to0}\frac{\frac2x}{-\frac2{x^3}}=\lim_{x\to0}-x^2=\boxed{0}$

7 0

3 years ago

Solve s = n/3 (b+H) for H Answer H=

cluponka [151]

The answer for H is, s=nb+nh÷3
Hope it helps

5 0

3 years ago