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Serggg [28]
3 years ago
15

Which of the following are considered measures of center? I. mean II. median III. interquartile range A. III only B. I and II on

ly C. I, II, and III D. I only
Mathematics
1 answer:
andrezito [222]3 years ago
4 0
Answer: Choice B) I and II only

The mean and median are considered measures of center as they represent the average of a data set. The average basically being a point that collectively speaks for all of the data. For example, if you have a group of basketball players whose heights range from 5'11" to 6'9", then there is no single height to report; however, we can compute the average to get a basic idea of the single height. The interquartile range (IQR) is a measure of variability or spread of the data. The higher the IQR, the more spread out the data is. Recall that 50% of the data is represented by the IQR and that IQR = Q3 - Q1 where Q1 and Q3 are the first and third quartiles respectively. So because the IQR is a measure of spread, it is not considered a center point.
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Step-by-step explanation:

When it comes to finding out how long it will take for an investment to double, one can use the Rule of 72.

The Rule of 72 estimates the amount of time it will take to double an investment when you divide 72 by the interest rate:

= 72 / r

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= 7.58 years

3 0
2 years ago
The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color bl
Alecsey [184]

Answer:

(a) The correct answer is P (CBM) = 0.79.

(b) The probability of selecting an American female who is not red-green color-blind is 0.996.

(c) The probability that neither are red-green color-blind is 0.9263.

(d) The probability that at least one of them is red-green color-blind is 0.0737.

Step-by-step explanation:

The variables CBM and CBW are denoted as the events that an American man or an American woman is colorblind, respectively.

It is provided that 79% of men and 0.4% of women are colorblind, i.e.

P (CBM) = 0.79

P (CBW) = 0.004

(a)

The probability of selecting an American male who is red-green color-blind is, 0.79.

Thus, the correct answer is P (CBM) = 0.79.

(b)

The probability of the complement of an event is the probability of that event not happening.

Then,

P(not CBW) = 1 - P(CBW)

                   = 1 - 0.004

                   = 0.996.

Thus, the probability of selecting an American female who is not red-green color-blind is 0.996.

(c)

The probability the woman is not colorblind is 0.996.

The probability that the man is  not color- blind is,

P(not CBM) = 1 - P(CBM)

                   = 1 - 0.004  

                   = 0.93.

The man and woman are selected independently.

Compute the probability that neither are red-green color-blind as follows:

P(\text{Neither is Colorblind}) = P(\text{not CBM}) \times  P(\text{not CBW})\\ = 0.93 \times  0.996 \\= 0.92628\\\approx 0.9263

Thus, the probability that neither are red-green color-blind is 0.9263.

(d)

It is provided that a one man and one woman are selected at random.

The event that “At least one is colorblind” is the complement of part (d) that “Neither is  Colorblind.”

Compute the probability that at least one of them is red-green color-blind as follows:

P (\text{At least one is Colorblind}) = 1 - P (\text{Neither is Colorblind})\\ = 1 - 0.9263 \\= 0.0737

Thus, the probability that at least one of them is red-green color-blind is 0.0737.

6 0
3 years ago
Reduce -7/8 + (-1/2) to the simplest form.
ololo11 [35]

Answer:

-1.375

Step-by-step explanation:

5 0
3 years ago
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