Answer:
0.14
Step-by-step explanation:
P(A|B) asks for the probability of A, given that B has happened. This is equal to the probability of A and B over the probability of B (see picture)
Here, the question is asking if someone is taking the bus given that they are a senior.
The probability of someone being a senior and taking the bus is 5/100, or 0.05 . The probability of someone being a senior is 35/100, or 0.35
Our answer is then 0.05/0.35 = 1/7 = 0.14
Answer:
Option C) The two quantities are equal.
Step-by-step explanation:
We are given the following in the question:
n is a negative integer and p is a positive integer.
Quantity A: The product of the integers from n to p
Quantity B: 0
Quantity A is the product of integers from n to p. Since all the integers from n to p will include zero between them, thus the product of all the integers from n to p will be zero.
The product takes place in the following manner:
Thus, both the quantities are equal.
Option C) The two quantities are equal.
Take the Laplace transform of both sides:
L[y'' - 4y' + 8y] = L[δ(t - 1)]
I'll denote the Laplace transform of y = y(t) by Y = Y(s). Solve for Y :
(s²Y - s y(0) - y'(0)) - 4 (sY - y(0)) + 8Y = exp(-s) L[δ(t)]
s²Y - 4sY + 8Y = exp(-s)
(s² - 4s + 8) Y = exp(-s)
Y = exp(-s) / (s² - 4s + 8)
and complete the square in the denominator,
Y = exp(-s) / ((s - 2)^2 + 4)
Recall that
L⁻¹[F(s - c)] = exp(ct) f(t)
In order to apply this property, we multiply Y by exp(2)/exp(2), so that
Y = exp(-2) • exp(-s) exp(2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-s + 2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-(s - 2)) / ((s - 2)² + 4)
Then taking the inverse transform, we have
L⁻¹[Y] = exp(-2) L⁻¹[exp(-(s - 2)) / ((s - 2)² + 4)]
L⁻¹[Y] = exp(-2) exp(2t) L⁻¹[exp(-s) / (s² + 4)]
L⁻¹[Y] = exp(2t - 2) L⁻¹[exp(-s) / (s² + 4)]
Next, we recall another property,
L⁻¹[exp(-cs) F(s)] = u(t - c) f(t - c)
where F is the Laplace transform of f, and u(t) is the unit step function
To apply this property, we first identify c = 1 and F(s) = 1/(s² + 4), whose inverse transform is
L⁻¹[F(s)] = 1/2 L⁻¹[2/(s² + 2²)] = 1/2 sin(2t)
Then we find
L⁻¹[Y] = exp(2t - 2) u(t - 1) • 1/2 sin(2 (t - 1))
and so we end up with
y = 1/2 exp(2t - 2) u(t - 1) sin(2t - 2)
