Answer:
The maximun height that the rocket will reach is 3150 ft.
Step-by-step explanation:
This problem requieres that we maximize the function. For this we need the function to be of the form:
We already have that covered, as the height of the rocket is represented by:
We start calculating the derivative of h(t), wich is:
We need to remember that the derivative of a function represents the slope of said function at a given point. The maximum value of the function will have a slope equal to zero.
So we find the value in wich the derivative equals zero:
This means that 15 seconds is the time at wich the rocket will reach it's maximum height.
We replace this value in the original equation to solve the problem:
With this we can conclude that the maximum height the rocket will reach is 3150 ft.
Answer:
a) 0.25
b) 0.38
c) 0.12
d) 0.64
Step-by-step explanation:
We have these following informations.
There are 500 people between 18 and 29. Of those 190 have at least one tattoo and 310 have no tattoo.
There are also 500 people between 30 and 50. Of those, 60 have at least one tattoo and 440 have no tattoo.
a) P(tattoo)
There are 1000 people. Of those, 190+60 = 250 have tattoo. So P(tattoo) = 250/1000 = 0.25.
b) P(tattoo|age 18-29)
There are 500 people aged 18-29. Of those, 190 have at least one tattoo. So P(tattoo|age 18-29) = 190/500 = 0.38.
c) P(tattoo|age 30-50)
There are 500 people aged 18-29. Of those, 60 have at least one tattoo. So P(tattoo|age 30-50) = 60/500 = 0.12.
d) P(age 18-29|tattoo)
250 people have at least one tattoo. Of those, 160 are between 18 and 29. So P(age 18-29|tattoo) = 160/250 = 0.64.
Answer:
a1= 1 q= −sinx , dla |q| <1 , ta suma jest zbieżna
a1 1
S=
=
1 −q 1+sinx
w mianowniku podobnie: a1=1 , q= sinx , dla | sinx| <1
1
S=
1 −sinx
i mamy równanie:
1
1+sinx
= tg2x
1
1− sinx
Step-by-step explanation:
No, because marathon runners are not a representative sample of all adults in the United States.
First, we find the time spent accelerating:
s = ut + 1/2 at²; u = 0
10 = 1/2 5t²
t = 2 seconds
Velocity at point B:
v = u + at; u = 0, a = 5 m/s², t = 2 s
v = (5)(2) = 10 m/s
Now, the same velocity is at C so we find the time decelerating to 0 in CD
v = u + at, v = 0, u = 10 m/s, a = -4 m/s²
0 = 10 - 4t
t = 2.5 seconds
Total time = time(AB) + time(BC) + time(CD)
20 = 2 + time(BC) + 2.5
time(BC) = 15.5 seconds
Distance = velocity x time
BC = 10 x 15.5
= 155 m