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FrozenT [24]
3 years ago
7

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were

admitted to their first choice college, as were 20% of the other students. You overhear a classmate say he got into the college he wanted. What is the probability he didn't take an SAT prep course?
What steps would I take to solve this problem?
Mathematics
1 answer:
Nesterboy [21]3 years ago
4 0

Answer: The probability he didn't take an SAT prep course = 0.985

Step-by-step explanation:

Let us first assume that he took SAT prep.

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were admitted to their first choice college. That is,

30/ 100 of 5 = 0.3 × 5 = 1.5

The probability he did take an SAT prep course and got admission into the college of first choice will be

P(prep) = 1.5 / 100 = 0.015

The probability he didn't take an SAT prep course will be:

P(not prep) = 1 - P(prep)

P(not prep) = 1 - 0.015

P(not prep) = 0.985

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what is the least number that must be added to 2000 so that the sum is divisible exactly 10,12,16and 18 ? do it step by step
Rudiy27

The integers divisible by any set of positive integers are the multiples of their LCM

let us first write the factored form of each

10 = 2×5

12 = 2×2×3

16 = 2×2×2×2

18 = 2 x3×3

Now we will find lcm of these numbers

LCM = 2×2×2×2×3×3×5 = 720

The multiples of 720 are divisible by 10,12,16 and 18.

2000/720 = 2.777777...

The least integer greater than that is 3, so 3×720 = 2160 is the least integer greater than 2000 that is divisible by 10,12,16 and 18.

so if we need to find what must be added to 2000 so that the sum is divisible by 10,12,16 and 18, we must subtract 2000 from 2160

2160-2000=160

so we must add 160 to 2000 so that the sum is divisible exactly 10,12,16and 18

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3 years ago
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Step-by-step explanation:

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3 years ago
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saveliy_v [14]

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Answer:

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Step-by-step explanation:

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