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3 years ago

Can u plz help me answer question 30

Mathematics

1 answer:

slamgirl [31]3 years ago

3 0

We are given:

$3x^3-2x^2$

There seems to be nothing I can do to simplify this equation. However, I can factor out an $x^2$ since they both have at least an $x^2$ . Here is what we get:

$x^2(3x-2)$

This is in the simplest form I can get into. I cannot do anything more to simplify this expression. This is your final answer.

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Using pemdas what’s the correct solution

SSSSS [86.1K]

4- 2/3 (4-1/6) divided by 3/4

parenthesis first

4 - 2/3 (3 5/6) divided by 3/4

change to an improper fraction (6*3+5)/6

4 - 2/3 ( 23/6)divided by 3/4

4 - 46/18 divide by3/4

copy dot flip

4 - 46/18 * 4/3

4 - 23/9 * 4/3

4 - 92/27

get a common denominator of 27

4*27/27 -92/27

108/27 - 92/27

16/27

4 0

3 years ago

Read 2 more answers

Select the correct answer.

ad-work [718]

Answer:

A' (-1, 2)

Step-by-step explanation:

(x, y) -> (-y, x)

A' (-1, 2)

B' (1, -2)

C' (2, -2)

D' (0, -2)

3 0

3 years ago

Express answer in exact form.

bazaltina [42]

The answer is 5.13 in²

Step 1. Calculate the diameter of the circle (d).
Step 2. Calculate the radius of the circle (r).
Step 3. Calculate the area of the circle (A1).
Step 4. Calculate the area of the square (A2).
Step 5. Calculate the difference between two areas (A1 - A2) and divide it by 4 (because there are total 4 segments) to get the area of one segment formed by a square with sides of 6" inscribed in a circle.

Step 1:
The diameter (d) of the circle is actually the diagonal (D) of the square inscribed in the circle. The diagonal (D) of the square with side a is:
D = a√2 (ratio of 1:1:√2 means side a : side a : diagonal D = 1 : 1 : √2)
If a = 6 in, then D = 6√2 in.
d = D = 6√2 in

Step 2.
The radius (r) of the circle is half of its diameter (d):
r = d/2 = 6√2 / 2 = 3√2 in

Step 3.
The area of the circle (A1) is:
A = π * r²
A = 3.14 * (3√2)² = 3.14 * 3² * (√2)² = 3.14 * 9 * 2 = 56.52 in²

Step 4.
The area of the square (A2) is:
A2 = a²
A2 = 6² = 36 in²

Step 5:
(A1 - A2)/4 = (56.52 - 36)/4 = 20.52/4 = 5.13 in²

6 0

3 years ago

Read 2 more answers

Solve −2x−4=3x+1 by graphing. The solution is x=

alex41 [277]

Answer: x=-1

Step-by-step explanation:

7 0

3 years ago

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

3 years ago