To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
<span>To do that, you need to set it all to zero and factor:
x^2+8x+15 = 0
(x+5)(x+3) = 0
then put both those in parentheses chunks in their own equations
x + 5 = 0
x + 3 = 0
and then simplify
x = -5
x = -3
So the two points where the parabola crosses the x-axis are -5 and -3.</span>
Answer:
0.2266
Step-by-step explanation:
We know that the grade point averages of a large population of college students are approximately normally distributed with a mean of 2.4 and a standard deviation of 0.8. The z-score related to 3.0 is computed as (3.0-2.4)/0.8 = 0.75. Therefore the probability that a randomly selected student will have a grade point average in excess of 3.0 is P(Z > 0.75) where Z comes from a standard normal distribution. So, P(Z > 0.75) = 0.2266
