All categories

3 years ago

2x+7 =5 is an example of...

Mathematics

1 answer:

tresset_1 [31]3 years ago

6 0

Answer:

This is an example of distribution

Step-by-step explanation:

The reason why is because the equation is linear and if you exclud the y it show that it can distribute.

You might be interested in

How do I solve for x in this equation?

Dafna11 [192]

\left[x \right] = \left[ 17\right][x]=[17]. totally answer

8 0

4 years ago

Peter's paint company made the same number of gallons of paint on Monday, Tuesday, and Wednesday. On Thursday, they made twice a

umka2103 [35]

Answer:

Peter's company uses on:

Monday = 4 gallons

Tuesday = 4 gallons

Wednesday = 4 gallons

Thursday = 8 gallons

Step-by-step explanation:

If they made a total of 20 gallons over the course of these four days, how many gallons did they make each day? Monday

Let the gallons of paint used be Peter's company on Monday, Tuesday, and Wednesday since it is the same amount be represented by x

On Thursday, they made twice as many as any of the previous days.

= 2x

Hence:

x + x + x + 2x = 20 gallons

5x = 20 gallons

x = 20 gallons/5

x = 4 gallons

Note that:

Thursday = 2x

Hence, = 2 × 4 gallons = 8 gallons

Hence, Peter's company uses on:

Monday = 4 gallons

Tuesday = 4 gallons

Wednesday = 4 gallons

Thursday = 8 gallons

5 0

3 years ago

What does statement that is not correct to use probability to describe an event.

loris [4]

Answer:BRAINLIEST PLEASE

Step-by-step explanation:

It is not possible right now and so if the probability is 0, then there wouldn't even be one.

8 0

3 years ago

Which of the following is a correct sine ratio for the figure? Question 20 options: A) C = 5∕12 B) C = 5∕13 C) C = 12∕13 D) C =

Andrei [34K]

Answer:

Step-by-step explanation:

Sine is opposite/hypotenuse. Sine C would be 5/13 so option B is correct.

3 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

4 years ago