The minor arc is equal to the measure of the _________

Nari says, "The sum of 10 and one-third of a number is 25." She uses the bar model to write an equation to represent her

Mariana [72]

Answer:

$25=10+\frac{1}{3}n$ .

Step-by-step explanation:

It is given that the sum of 10 and one-third of a number is 25.

Let as consider n be the unknown number.

One-third of a number $=\frac{1}{3}n$

Sum of 10 and one-third of a number $=10+\frac{1}{3}n$

The sum of 10 and one-third of a number is 25. It can be written as

$25=10+\frac{1}{3}n$

Therefore, the required equation is $25=10+\frac{1}{3}n$ .

4 0

2 years ago

Read 2 more answers

At a grocery store, an uncooked beef roast is on sale for $4.99/lb. At the same store, prepared roast beef is at the deli for $2

Mrrafil [7]

For us to find out how many times more expensive is the deli roast compared to the uncooked roast we need to change the units;
cost of 100g=0.221lb is $2.99
cost of 1 lb will therefore be:
1/0.221*2.99
=$13.53/lb
therefore the number of times more expensive the deli roast is compared to uncooked roast is:
[price of 1lb roasted meat]/[price of 1 lb uncooked meat]
=13.53/4.99
=2.7 times

6 0

2 years ago

Calculate the correlation coefficient for the given data below:

Rasek [7]

The correlation coefficient of the data provided is -0.52.

<h3>What is the correlation coefficient?</h3>

Correlation is a measure that is used in statistics to determine the linear relationship that exists between two variables. Correlation can be positive, negative or zero.

Positive correlation is when two variables move in the same direction. Negative correlation is when two variables move in opposite directions. Zero correlation is when there is no relationship between the variables.

Correlation can be determined using a financial calculator. When the financial calculator is used, the correlation coefficient is -0.52

To learn more about negative correlation, please check: brainly.com/question/27246345

#SPJ1

4 0

10 months 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

Suppose you are interested in using regression analysis to estimate an NBA player's salary using the following independent varia

lions [1.4K]

Answer:

a. the player was traded in the last 5 years

4 0

2 years ago

The minor arc is equal to the measure of the __________