Select all that apply which expressions are equivalent to 6X +12
1 answer:
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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)
Average reading score is given by the expression,
y = 0.153x + 255.4
Here, x in the number of years past 1970.
To find the year in which the average reading score is 259.378, substitute the value of 'y' in the given expression.
y = 0.153x + 255.4
259.378 = 0.153(x) + 255.4
0.153x = 259.378 - 255.4
0.153x = 3.978
x =
x = 26
That means given average reading score 259.378 will be 26 years after 1970.
Therefore, 1996 is the year in which the average reading score will be 259.378.
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Answer:
Step-by-step explanation:
Let us record the station number 1, 2 or 3 for each family member A, B or C.
I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.
From table it is clear that the total outcomes possible are 27.
We know that, probability can be defined as,
a) All Members Assigned to the Same Station.
Cases for all members being assigned to same station are as follows:
[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).
Therefore,
b) At Most Two Members Assigned to the Same Station.
It means that maximum of 2 members can have the same station. Cases for this situation are as follows:
[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],
[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]
(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).
Therefore,
c) All Members Assigned to a Different Station.
For this scenario, we have the following results:
[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).
Therefore,
