50%. students (in percent) who passed the second exam also passed the first exam.
Let's imagine that there are 100 kids in the teacher's class. We know that 40 of them passed BOTH tests, and 80 passed the second test.
Because if they weren't, they wouldn't have passed the first test and consequently wouldn't have passed both, we can be sure that the group of students who passed BOTH tests is only made up of the 80 who passed the second test.
Thus, both tests were passed by 40 of the 80 pupils who passed the second one:
40/80 = 1/2 = 50%.
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Answer:
- 3n+6 (n = smallest)
- 3n (n = middle)
Step-by-step explanation:
The usual method for doing this is to let n represent the smallest one. Then the three integers are ...
n, n+2, and n+4
and their sum is ...
(n) +(n+2) +(n+4) = 3n+6 . . . . sum of 3 consecutive odd integers (n = smallest)
_____
Personally, for consecutive number problems, I prefer to let the variable represent the average value. If n is the average value of 3 consecutive odd integers, is is the middle integer. Of course, the sum will be 3 times the average:
(n-2) +(n) +(n+2) = 3n . . . . sum of 3 consecutive odd integers (n = middle one)
I think Ms.Gonzalez got 217 prizes on Wednesday.All you had to do was divide 1,307 by 6.
Answer:
C. (0,b)
Step-by-step explanation:
Answer:
B. -1
Step-by-step explanation:
examples:
1 and -1
1/2 and -2
