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 + 128t. What is the interval of time during which the projectile's height exceeds 192 feet?

  (2, 6)

Step-by-step explanation:

We want h > 192, so ...

  -16t^2 +128t > 192

  t^2 -8t < -12 . . . . . . . divide by -16

  t^2 -8t +12 < 0 . . . . . add 12

  (t -6)(t -2) < 0 . . . . . . factor

The zeros of this are t=2 and t=6. The signs of the factors will be different, hence their product will be negative, for values of t between these zeros.

The projectile's height exceeds 192 feet in the interval 2 < t < 6.

c

step-by-step explanation:

because of the levels

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts