The probability that a data-communications system will have high selectivity is .72, the probability that it will have high fidelity is .59, and the probability that it will have both is .33. Find the probability that a system with high fidelity will also have high selectivity.

The probability that a system with high fidelity will also have high selectivity is 0.559322

Step-by-step explanation:

According to the given data we have the following:

P(high selectivity)=0.72

P(high fidelity )=0.59

P(both)=0.33

Therefore, to calculate the probability that a system with high fidelity will also have high selectivity we would have to make the following calculation:

the probability is  P(high selectivity|high fidelity)  =P(both)/P(high fidelity)

the probability is  P(high selectivity|high fidelity) =0.33/0.59

the probability is  P(high selectivity|high fidelity) = 0.559322

The probability that a system with high fidelity will also have high selectivity is 0.559322

80%

step-by-step explanation:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: