I need help what is 2|5 / -2

What is the y-intercept of the line with the equation 4x + 2y=12

malfutka [58]

The y-intercept (0,6) all you have to do when the formula is written like that just divide the opposite side by the number next to y.

3 0

3 years ago

Read 2 more answers

Just say the answer pls and ty

Nonamiya [84]

Answer:

the answer pls and ty

7 0

3 years ago

ms.marks records the number of words each typist can type in 1 minute how many more words could the fastest typist type in 30 mi

krok68 [10]

The fastest typist is Charlie at 60wpm. Let's figure out how many words he can type in 30 minutes using place value strategies.

6 * 3 = 18

Now, move the decimal two places to the right because there is one zero in 60 and one zero in 30.

1800.

Charlie can type 1800 words in 30 minutes.

The slowest typist is Jerome, at 40 wpm. Let's figure out how many words he can type in 30 minutes using place value strategies.

4 * 3 = 12

Now, move the decimal two places to the right because there is one zero in 40 and one zero in 30.

1200.

Jerome can type 1200 words in 30 minutes.

Now, subtract Charlie's words in 30 min by Jerome's words in 30 min.

1800 - 1200 = 600

Charlie can type 600 more words in 30 minutes than Jerome.

7 0

3 years ago

Please help me! I will give a lot of points and brainliest if correct!

Snezhnost [94]

Answer:

The answer is B

Step-by-step explanation:

A farm is in the shape of a rectangle, 1/8 of a mile long and 2/5 of a mile wide. What is the area of the farm?

When looking for area, you will always multiply or divide.

The multiplication in this problem is 2/5 x 1/8, which is the same as the Multiplication Operation given.

So, the answer is B.

6 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

I need help what is 2|5 / -2​

I need help what is 2|5 / -2