Multiply. Write your answer in simplest form. 5/7 x 7/10

When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode

yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are central tendency measures in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a normal distribution, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a distribution is not symmetrical, we say it is skewed. The skewness is a measure of the asymmetry of the distribution. In this case, the mean, median and mode are not the same, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (negative skew), or greater than them (positive skew), or approximately equal than the median but much greater than the mode (a variation of a positive skew case).

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a mound-shaped symmetrical distribution, it resembles the normal distribution and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, they are all approximately equal. So, the general relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, considering they have similar characteristics of the normal distribution, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

5 0

3 years ago

As a motivational speaker, Bree speaks at school assemblies, charging a school

Oksi-84 [34.3K]

Answer:

Bree spoke at 55 assemblies in this district.

Step-by-step explanation:

Bree speaks at school assemblies charging a school district a total cost C modeled by the equation:

C(a) = 50a+1,500

Being a the number of assemblies attending the meeting.

It's given Bree charged the Escambia City School District $4,250 to speak at school assemblies.

We can find the number of assemblies by solving the equation:

50a+1,500=4,250

Subtracting 1,500:

50a=4,250-1,500=2,750

Dividing by 50;

a = 2,750 / 50=55

Bree spoke at 55 assemblies in this district.

4 0

3 years ago

7.

Serjik [45]

Answer:

Probability that a customer pays the bill by mail, given that the customer is female is 0.20.

Step-by-step explanation:

The complete question is:

Online By Mail Total

Male 12 8 20

Female 24 6 30

Total 36 14 50

A utility company asked 50 of its customers whether they pay their bills online or by mail. What is the probability that a customer pays the bill by mail, given that the customer is female.

Probability that a customer pays the bill by mail, given that the customer is female is given by = P(Pays bill by mail / Customer is female)

$\frac{6}{50}$ P(Pays bill by mail/Customer is female) = $\frac{P(\text{Pays bill by Mail}\bigcap \text{ Customer is Female})}{P(\text{ Customer is Female})}$