Answer:

Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
Step-by-step explanation:
Normal with mean 6.88 minutes and
a standard deviation of 0.57 minutes.
Choose a student at random from this group and call his time for the mile Y. Find P(Y<6)


y ≈ normal (μ, σ)
The z score is the value decreased by the mean divided by the standard deviation

Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
F(x) = (x + 1)(x - 2)
f(x) = x(x - 2) + 1(x - 2)
f(x) = x(x) - x(2) + 1(x) - 1(2)
f(x) = x² - 2x + x - 2
f(x) = x² - x - 2
It is
(3-x)^2
i made sure it correct this time
So there where two times the amount of rock tickets sold than the jazz concert. If that is it then you divide the 1840 by 2 getting 920, dividing 1÷2 you get zero remainder 1, 18÷2= 9, 4÷2= 2, and 0÷2= 0. 920. Hope i helped :)
<span>22 + (3^2 – 4^2)
= </span><span>22 + (9 – 16)
</span><span>= 22 -7
= 15</span>