If p^2 – 13p + 40 = s, and p is a positive integer between 1 and 10, inclusive, what is the ...
If p^2 – 13p + 40 = s, and p is a positive integer between 1 and 10, inclusive, what is the probability that s < 0?
Answer/Solution
1/5
Steps/Work
p2 – 13p + 40 = s
so (p – 8)(p – 5) = s
For q to be negative, the expressions (p – 8) and (p – 5) must have opposite signs. Which integers on the number line will yield opposite signs for the expressions (p – 8) and (p – 5)? Those integers in the range 5 < p < 8 (notice 5 and 8 are not included because they would both yield a value of zero and zero is a nonnegative integer). That means that there are only two integer values for p, 6 and 7, that would yield a negative q. With a total of 10 possible p values, only 2 yield a negative q, so the probability is 2/10 or 1/5.
The correct answer is B.
so (p – 8)(p – 5) = s
For q to be negative, the expressions (p – 8) and (p – 5) must have opposite signs. Which integers on the number line will yield opposite signs for the expressions (p – 8) and (p – 5)? Those integers in the range 5 < p < 8 (notice 5 and 8 are not included because they would both yield a value of zero and zero is a nonnegative integer). That means that there are only two integer values for p, 6 and 7, that would yield a negative q. With a total of 10 possible p values, only 2 yield a negative q, so the probability is 2/10 or 1/5.
The correct answer is B.