Simplify. Write answers with positive exponents only.

Mathematics

1 answer:

allochka39001 [22]3 years ago

7 0

Answer:

The answer to your question is 2⁸x²⁰ y²⁴ or 256x²⁰y²⁴

Step-by-step explanation:

Data

(2x³y³)⁴ (2x²y³)⁴

To solve this problem, use the law of exponents.

1.- Use the law of Power of a power rule

(2x³y³)⁴ (2x²y³)⁴ = (2⁴x¹²y¹²) (2⁴x⁸y¹²)

2.- Use the multiplication rule

(2⁴x¹²y¹²) (2⁴x⁸y¹²) = 2⁸x²⁰ y²⁴

3.- The answer is 2⁸x²⁰y²⁴ or we can simplify more like

256x²⁰y²⁴

You might be interested in

An international company has 13,600 employees in one country. If this represents 19.3% of the company's employees, how many empl

user100 [1]

Hi there! To solve this question, we can write and solve a proportion. Let’s set it up like this: 13,600/x = 19.3/100. We do this, because we’re looking for the total amount of people, which would be the whole. We already have the part. Cross multiply the values in order to get 1,360,000 = 19.3x. Now, divide each side by 19.3 to isolate the variable. When you do, you get x is approx. 70,466.3212 or 70,466 when rounded to the nearest whole number. Plus, you can’t have a fraction of a person, anyway. The company has a total of 70,466 employees.

4 0

3 years ago

A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi

34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{X}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{X}=\sqrt{n}\sigma$

It is provided that:

$\mu=\$27\\\sigma=\$18\\n=310$

As the sample size is quite large, i.e. n = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

$P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z$