Writer and solve a real-word promblem involving the multiplication of a fraction and a whole number whose is between 10and15

Which of the following equations is equivalent to 3x + 4y =15

slavikrds [6]

Answer:

x= −4 /3 y+5

Step-by-step explanation:

5 0

3 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

2 years ago

How to do you find the constant of variation when it’s a direct variation?

ololo11 [35]

Answer:

The constant of variation in a direct variation is the constant (unchanged) ratio of two variable quantities. The formula for direct variation is. y = k x (or y = k x ) where k is the constant of variation . Example 1: If y varies directly as x and y = 15 when x = 24 , find x when y = 25 .

Step-by-step explanation:

Hope this helps, mark brainliest

3 0

2 years ago

Combine Like Terms:<br><br> 7X + 3M - 2(2X + 8) + 24 + 8M - X<br> so hard

Elza [17]

Answer:

11M + 2X + 8

Step-by-step explanation:

Hope this helps!! :))

7 0

2 years ago

Read 2 more answers

What are the relative frequencies, to the nearest hundredth, of the rows of the two way table?

Nookie1986 [14]

Answer: A B

Group A 0.25 0.75

Group B 0.45 0.55

Step-by-step explanation:

Since we have given that

Number of people of A in Group A = 15

Number of people of B in Group A = 45

Number of people of A in Group B = 20

Number of people of B in Group B = 25

Total number of people in Group A = 15+45 = 60

Total number of people in Group B = 20+25 = 45

So, Relative frequencies will be given below:

Relative frequency of A in Group A is given by

$\frac{15}{60}=0.25$

Relative frequency of B in Group A is given by

$\frac{45}{60}=0.75$

Relative frequency of A in Group B is given by

$\frac{20}{45}=0.45$

Relative frequency of B in Group B is given by

$\frac{25}{45}=0.55$

Hence, A B

Group A 0.25 0.75

Group B 0.45 0.55

5 0

3 years ago