3a+ 6 = 15, 3a = 9, a+ 2= 5, 1/3a =1
1 answer:
Given parameters:
Find if any of the equations are the equivalent;
3a+ 6 = 15
3a = 9
a+ 2= 5
=1
To find which of the equations are equivalent to one another, let us solve them first. The ones with the same solution are definitely equivalent.
(i) 3a+ 6 = 15
3a = 15 - 6
3a = 9
a = 3
(II) 3a = 9
a = 3
(III) a+ 2= 5
a = 5-2
a = 3
(IV) =1
1 = 3a
a =
We clearly see that the first three equations are equivalent.
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Answer:
c = 0.165
Step-by-step explanation:
Given:
f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,
f(x, y) = 0 otherwise.
Required:
The value of c
To find the value of c, we make use of the property of a joint probability distribution function which states that
where a and b represent -infinity to +infinity (in other words, the bound of the distribution)
By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have
Since c is a constant, we can bring it out of the integral sign; to give us
Open the bracket
Integrate with respect to y
Substitute 0 and 3 for y
Add fraction
Rewrite;
The is a constant, so it can be removed from the integral sign to give
Integrate with respect to x
Substitute 0 and 3 for x
Multiply both sides by