The R. R. Bowker Company of New York collects data on annual subscription rates to periodicals. Results are published in Library Journal. In a recent independent study, it was found that 63% of all students at Mill University read Time magazine, 51% read U. S. News and World Report, and 24% read both magazines. If a student at Mill University is randomly selected, what is the probability that the student reads either the Time magazine or the U. S.News and World Report magazine?

90%

Step-by-step explanation:

Let's call the percentage of students that read Time magazine by P(T), and the percentage of students that read U.S News and World Report by P(U). So, we have that:

P(T) = 0.63

P(U) = 0.51

P(T and U) = 0.24

To find the percentage of students that read either the Time magazine or the U.S.News and World Report magazine (that is, P(T or U)), we can use this formula:

P(T or U) = P(T) + P(U) - P(T and U)

So, we have that:

P(T or U) = 0.63 + 0.51 - 0.24 = 0.90

So the probability is 90%

domain is all the x values included in the

0 ≤ x ≤ 2

range is all the y values of the graph

-2 ≤ y ≤ 2

***keep in mind that there are different ways to show domain and range

hope this !

~just a girl in love with shawn mendes

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg