Answer: I think it would be a 50/50 chance of getting 2 even numbers.
Answer:
Kevin needs 96 points on his last test to raise his mean test score to 90 points.
Step-by-step explanation:
we know that
The mean score is the total of all scores divided by the total number of tests.
Let
x_1 ----> the score in the first math test
x_2 ----> the score in the second math test
x_3 ----> the score in the third math test
x_4 ----> the score in the fourth math test
we have
After taking the first 3 tests, his mean test score is 88 points
so

----> equation A
How many points does he need on his last test to raise his mean test score to 90 points?
so

----> equation B
substitute equation A in equation B

solve for x_4


Therefore
Kevin needs 96 points on his last test to raise his mean test score to 90 points.
When you have a coefficient in front of the "x^2" value, you have to multiply it by the end value (4)
so you want a set of integers that will add up to 29, by multiple to (7 x 4) = 28
The only numbers like that are 28 and 1, they add up to 29 and multiply to 29
Therefore, rewrite it like this 7x^2 + 28x + x + 4
and then factor out the 7x in the first two terms, to get this
7x(x+4) + x + 4, then factor out 1 in the latter two terms to get this:
7x(x+4) +1(x+4), then use grouping to combine what you've got:
The factors are: {{ (7x+1)(x+4) }}
3(5)+(-2+4(5))
15+(-2+20)
15+(18)
15+18
=33
Do you have answer choices?