Answer:
Let X = score of final exam.
X~normal(23, 16)
(a)
percentage of score between 19 and 27
= P(19 < X < 27)
= P((19-23) / sqrt(16) < Z < (27 - 23) / sqrt(16))
= P(-1 < Z < 1)
= 1 - P(Z <= -1) - P(Z >= 1)
= 1 - P(Z >= 1) - P(Z >= 1)
= 1 - 2*P(Z >= 1)
= 1 - 2(0.1587)
= 0.6826
= 68.26%
According to Normal Distribution Table
P(Z>=1) = 1 - P(Z<1) = 1-0.8413
So the final percentage is 68.26%
Step-by-step explanation:
I'm not sure but I think there are 2 terms!
Answer:
It’s the third choice : x ≥8
Step-by-step explanation:
the third choice means he made equal to or greater than $8 an hour.
185-25=160
160÷20=8
Answer:
A.
Step-by-step explanation:
so, the equation is
h(t) = -t² + 7t
so, we need to find the solutions for t (the time when the ball is exactly 10 ft in the air). there had to be 2 solutions, as the ball first goes up passing the 10 ft height, and then comes back down again, passing the 10 ft mark a second time. and between these 2 times the ball is higher (but not equal, so, we can only use < or > as inequality signs) than 10 ft.
10 = -t² + 7t
-t² + 7t - 10 = 0
the generation solution to a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = t
a = -1
b = 7
c = -10
t = (-7 ± sqrt(7² - 4×-1×-10))/(2×-1) =
= (-7 ± sqrt(49 - 40))/-2 = (-7 ± sqrt(9))/-2
t1 = (-7 + 3)/-2 = -4/-2 = 2 seconds
t2 = (-7 - 3)/-2 = -10/-2 = 5 seconds
so, between 2 and 5 seconds airtime the ball is higher than 10 ft.
and remember : HIGHER THAN.
so, we cannot use any equality (like <= or >=).
t must be higher than 2 and lower than 5 :
2 < t < 5
A rational number can be expressed in the form a/b, where a and b are other integers. To satisfy this definition, 0.9 can be written as 9/10, 18/20, 90/100, etc.
