Will give brainliest! A survey of 1000 students found that 41% of the surveyed students were freshmen, 31% were juniors, and 67% lived on the college campus. (A) If 260 of the surveyed students were freshmen living on the college cam- pus, then how many surveyed students were either freshmen or living on campus? (B) Suppose from a previous survey you know that of interviewed juniors, only 23% live on campus. What is the probability a surveyed student is a junior or someone that lives on campus?

(A) 560 students

(B) P = 0.41

Step-by-step explanation:

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As given, a survey of 1000 students found that:

41% of the surveyed students were freshmen

=> 1000 x 41/100 = 410 surveyed students were freshmen

31% were juniors

=> 1000 x 31/100 = 310 surveyed students were junior

67% of surveyed students lived on the college campus

=> 1000 x 67/100 = 670 surveyed students lived on the campus

(A) 260 of the surveyed students were freshmen living on the campus

=> A = 410 - 260 = 150 surveyed students were freshmen NOT living on the campus

=> B = 670 - 260 = 410 surveyed students were not freshmen living on the campus

=> The number of surveyed students who were either freshmen or living on the campus:

N = A + B = 150 + 410 = 560

(B)  From a previous survey you know that of interviewed juniors, only 23% live on campus (we don't really need this)

=>The probability a surveyed student is a junior or someone that lives on campus: P = (670 - 260)/1000 = 0.41

I hope this helps!

answer:

what is your question that you ask

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