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3 years ago

Q1:

2 answers:

8 0

I only did Q1: No, the are not equal. X^3*x^3*x^3= x^9 and x^3*3*3=9x^3. They are not equal because when you multiply exponents by exponents, it’s like you are adding them.

Does this answer your question?

Shtirlitz [24]3 years ago

7 0

Answer:

D Now Mark Me Brainliest Please

Step-by-step explanation:

You might be interested in

3x+9y=12 <br>-3x-6y=18 <br>Solve the system by using elimination

Ivenika [448]

Answer: (x, y) = (-26, 10)

This means x = -26 and y = 10 pair up together

===============================================================

How to get that answer:

Add the equations straight down. Think of having 3 separate columns for the three different like terms.

3x + (-3x) becomes 0x and that's just 0. So the x terms go away
9y + (-6y) is the same as 9y-6y which turns into 3y
12 + 18 becomes 30

After all that, we're left with the simpler equation 3y = 30. That solves to y = 10 after dividing both sides by 3.

Plug this back into any equation involving x and y. Solve for x

3x+9y = 12

3x+9(10) = 12

3x+90 = 12

3x = 12-90

3x = -78

x = -78/3

x = -26

We have found that x = -26 and y = 10 pair up together to form the solution.

-------------------

Checking the answer:

Plug (x,y) = (-26, 10) into the first equation

3x+9y = 12

3(-26)+9(10) = 12

-78+90 = 12

12 = 12

We get a true equation since the same thing is on both sides.

Repeat the same idea for the second equation

-3x-6y = 18

-3(-26) - 6(10) = 18

78 - 60 = 18

18 = 18

The second equation is true as well.

Both equations are true, so the solution (x, y) = (-26, 10) is confirmed.

8 0

3 years ago

Read 2 more answers

A catering company provides packages for weddings and for showers. The cost per person for small groups

Tomtit [17]

Using the <em>normal distribution and the central limit theorem</em>, it is found that the probability the mean cost of the weddings is more than the mean cost of the showers is of 0.9665.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

When two variables are subtracted, the mean is the subtraction of the means, while the standard error is the square root of the sum of the variances.

<h3>What is the mean and the standard error of the distribution of differences?</h3>

For each sample, they are given by:

$\mu_W = 82.3, s_W = \frac{18.2}{\sqrt{9}} = 6.0667$

$\mu_S = 65, s_S = \frac{17.73}{\sqrt{6}} = 7.2382$

For the distribution of differences, we have that:

$\mu = \mu_W - \mu_S = 82.3 - 65 = 17.3$

$s = \sqrt{s_W^2 + s_S^2} = \sqrt{6.0667^2 + 7.2382^2} = 9.4444$

The probability the mean cost of the weddings is more than the mean cost of the showers is P(X > 0), that is, <u>one subtracted by the p-value of Z when X = 0</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$