Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
The answer is 18125
explanation:
rewrite the equation set = to 0.
x^2 + 5x - 8 = 0
The quadratic will not factor so you have to use the quadratic formula.
x = (-b + - sqrt(b^2 - 4ac))/2a
x = (5 + - sqrt(25 - 4* 1* -8))/2
x = (5 + - sqrt 57)/2
The x2is not the same as 2x. It is x^2. X tot he second power which makes the problem a quadratic equation. You cannot combine the terms x^2 and -5x because they so not have the same power.
Answer:
C) 3,-2,-7,-12
Step-by-step explanation:
A sequence is defined recursively using the formula f(n+1)=f(n)-5. Therefore, 3,-2,-7,-12 is the sequence that could be generated using this formula.
