To solve this problem, we
have to use the formula:
P = 1 - e^(-λx)
<span>where λ = 0.01382, and x
is the distance</span>
A. when x = 100
(P < or equal to 100) =
1 - e^(-0.01382 * 100)
(P < or equal to 100) =
0.75
B. when x = 200
<span>(P < or equal to 200) = 1 - e^(-0.01382 * 200)
(P < or equal to 200) = 0.94</span>
C. when 100 < x < 200
P (100 < x < 200) = P ((x < 100) + P(x > 200))
P (100 < x < 200) = 0.75 + (1 – 0.94)
<span>P (100 < x < 200) =
0.81 </span>
Answer: C.) a and d
Step-by-step explanation:
The equation modeling the height of the ball at time,t is given by h=-16t²+48t+6
<span>a. In how many seconds does the ball reach its maximum height?
Time taken for the ball to reach the maximum height will be given by:
h'(t)=-32t+48=0
finding the value of t we get:
32t=48
t=48/32
t=1.5 seconds
Thus the time taken for the ball to reach the maximum time is 1.5 seconds
b]</span><span> What is the ball’s maximum height?
The maximum height will be:
h(1.5)=-16(1.5)</span>²+48(1.5)+6
h(1.5)=42 fee
Answer:
B
Step-by-step explanation:
Answer:
277
Step-by-step explanation:
PEMDAS
E: 3^3 (3x3x3) = 27
M: 27x10 = 270
A: 270+7 = 277
