They all add up to 2.5 maybe u have to simplify?
Answer:
Probability at least one car will get punctured: 0.39347
Step-by-step explanation:
B(10,000 , 0.00005)
P(X ≥ 1) = 1 - P(X = 0)
= 1 - (1 - 0.00005)^10,000
= 1 - (0.99995)^10,000
= 1 - 0.60652...
= 0.39347 (probability that at least one car will get punctured)
As you can tell P(X ≥ 1) as we have to solve for the probability that at least one car will get punctured. That is of course 1 - [ P(X = 0) ].
(a) The "average value" of a function over an interval [a,b] is defined to be
(1/(b-a)) times the integral of f from the limits x= a to x = b.
Now S = 200(5 - 9/(2+t))
The average value of S during the first year (from t = 0 months to t = 12 months) is then:
(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12
or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12
This equals 200/12 * (5t -9ln(2+t))
Evaluating this with the limits t= 0 to t = 12 gives:
708.113 units., which is the average value of S(t) during the first year.
(b). We need to find S'(t), and then equate this with the average value.
Now S'(t) = 1800/(t+2)^2
So you're left with solving 1800/(t+2)^2 = 708.113
<span>I'll leave that to you</span>
Answer:
f(- 4) = 12
Step-by-step explanation:
To evaluate f(- 4) substitute x = - 4 into f(x) , that is
f(- 4) = - 8 - 5(- 4) = - 8 + 20 = 12
Only 1 and 51 are factors of 51.
