Let "a" and "s" represent the costs of advance and same-day tickets, respectively. Your problem statement gives you two relations.
.. a + s = 35 . . . . . the combined cost of one of each is 35
.. 15a +40s = 900 . . total paid for this combination of tickets was 900
There are many ways to solve these equations. You've probably been introduced to "substitution" and "elimination" (or "addition"). Using substitution for "a", we have
.. a = 35 -s
.. 15(35 -s) +40s = 900 . . substitute for "a"
.. 25s +525 = 900 . . . . . . . simplify
.. 25s = 375 . . . . . . . . . . . .subtract 525
.. s = 15 . . . . . . . . . . . . . . .divide by 25
Then
.. a = 35 -15 = 20
The price of an advance ticket was 20.
The price of a same-day ticket was 15.
Should be the last one, 43, 52, 61.
Answer:
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Step-by-step explanation:
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Answer: 35
Step-by-step explanation:
Given : The IQs of 700 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 11.
i.e.
and 
Let x denotes the IQs of applicants to college.
If the college requires an IQ of at least 97, then, the probability that students have IQ less than 97:-
[By using z-table]
Number of students will be rejected on this basis of IQ = Total students x Probability of students have IQ less than 97
= 700 x 0.0505 = 35.35 ≈ 35
Hence, about 35 students will be rejected on this basis of IQ, regardless of their other qualifications .