A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet? A. 3 B. t<4
C. t>4
D. 3>t>4

3 < t < 4

Step-by-step explanation:

Given the following information :

Initial Velocity of projectile = 112 ft/s

Height (h) above the ground after t seconds :

h = –16t2 + 112t

To calculate the time when h exceeds 192 Feets (h >192)

That is ;

-16t^2 + 112t = > 192

-16t^2 + 112t - 192 = 0

Divide through by - 16

t^2 - 7t + 12 = 0

Factorizing

t^2 - 3t - 4t + 12 =0

t(t - 3) - 4(t - 3) = 0

(t-3) = 0 or (t-4) =0

t = 3 or t=4

Therefore, t exist between 3 and 4 for height in excess of 192ft

–16(3)^2 + 112(3) = 192 feets

–16(4)^2 + 112(4) = 192 feets

3 < t < 4

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.