Answer:
a) 72.25sec
b) 6.25secs
c) after 10.5secs and 2 secs
Step-by-step explanation:
Given the height reached by the rocket expressed as;
s(t)= -4t^2 + 50t - 84
At maximum height, the velocity of the rocket is zero i.e ds/dt = 0
ds/dt = -8t + 50
0 = -8t + 50
8t = 50
t = 50/8
t = 6.25secs
Hence it will reach the maximum height after 6.25secs
To get the maximum height, you will substitute t - 6.25s into the given expression
s(t)= -4t^2 + 50t - 84
s(6.25) = -4(6.25)^2 + 50(6.25) - 84
s(6.25) = -156.25 + 312.5 - 84
s(6.25) = 72.25feet
Hence the maximum height reached by the rocket is 72.25feet
The rocket will reach the ground when s(t) = 0
Substitute into the expression
s(t)= -4t^2 + 50t - 84
0 = -4t^2 + 50t - 84
4t^2 - 50t + 84 = 0
2t^2 - 25t + 42 = 0
2t^2 - 4t - 21t + 42 = 0
2t(t-2)-21(t-2) = 0
(2t - 21) (t - 2) = 0
2t - 21 = 0 and t - 2 = 0
2t = 21 and t = 2
t = 10.5 and 2
Hence the time the rocket will reach the ground are after 10.5secs and 2 secs
1. 21/4 , you can use a calculator it’s way easier because your doing times
Answer:
3/11 divided by 3/11 is 1
9/10 divided by 3/5 is 1 1/2 (1.5)
Step-by-step explanation:
Answer:
73.6
61.824
.2776
2.3
1.342
.4129
*note* I had trouble reading the standard deviation of the second part, so if it's not 3 then leave a comment and i will fix it*
Step-by-step explanation:
1.)
the mean would be something like 460*.16= 73.6
The variance should be 460*.16(1-.16)= 61.824 which means the standard deviation is √61.824= 7.863
for p to be less than .15 it would have to be less than .15*460= 69
(69-73.6)/7.863= -.59 which has a probability of .2776
2.)
the mean is 2.3
I am not sure if the standard deviation is 3 or .3 (picture is kind of blurry)
I am thinking that it's three which would mean that the standard deviation would be √(3²*(1/5))= 1.342
(2-2.3)/1.342= -.22 = .4129
