Answer:
a. Expected frequency of losses per semester = 0*0.08 + 5*0.16 + 10*0.28 + 15*0.32 + 20*0.14 + 25*0.02
Expected frequency of losses per semester = 11.7 losses per semester
b. Variance = (0-11.7)²*0.08 + (5-11.7)²*016 + (10-11.7)²*0.28 + (15-11.7)²*0.32 + (20-11.7)²*0.14 + (15-11.7)²*0.02
Variance = 10.9512 + 7.1824 + 0.8092 + 3.7848 + 9.6446 + 3.5378
Variance = 35.61
c. As losses equal $60, expected losses per semester = 11.7*$60 = $702
d. Expected losses of all textbooks per semester = 250*11.7*$60 = $175,500
Answer:
Two solutions
Step-by-step explanation:
This is a quadratic equation in the form y = ax² + bx + c.
For quadratic equations, you can find solutions using the quadratic formula:
.
To find the number of solutions, <u>you only need what's inside the square root</u>. We call it the "<u>discriminant</u>" because lets us know the number of solutions without solving.

If b²- 4ac > 0, two solutions. (greater than)
If b²- 4ac < 0, no solutions. (less than)
If b²- 4ac = 0, one solution. (equal to)
y = ax² + bx + c
y = -3x² + x + 12
a = -3 b = 1 c = 12
<u>Substitute into the discriminant</u>
b²- 4ac
= 1² - 4(-3)(12)
= 1 - (-144)
= 145 > 0
b²- 4ac > 0 Discriminant greater than 0
Therefore, there are two solutions.
30 - 2(7 + 2) - 1
Distribute -2 into the parenthesis:
30 - 14 - 4 - 1
Subtract from left to right:
16 - 4 - 1
12 - 1
11