The 100-meter race times at a state track meet are normally distributed with a mean of 14.62 seconds and a standard deviation of 2.13 seconds. using the standard normal probabilities table, what is the approximate probability that a runner chosen at random will have a 100-meter time less than 15.5 seconds? a.0.1894 b. 0.3409 c. 0.6591 d. 0.7910 e. 0.8106

To solve this problem, we use the t statistic. The formula for z score is:

z = (x – u) / s

where x is the sample value = 15.5 seconds or below, u is the sample mean = 14.62 seconds, s is standard dev = 2.13 seconds

z = (15.5 – 14.62) / 2.13

z = 0.41

Using standard distribution tables at z = 0.41, the value of P is:

P = 0.6591 = 65.91%

Hence there is a 65.91% chance the runner will have less than 15.5 seconds of time

c. 0.6591

lol we need to see what the question is asking about

step-by-step explanation:

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.