Why might the mean of a probability distribution for a discrete random variable be less than (or greater than) the average of po
1 answer:
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Here it is given that the width is x ft and total length of the fence is 2400 ft .
Let the length be y ft
So we have
Let A represents area, and area is the product of length and width .
So we get
Substituting the value of y, we will get
Second part
The area is maximum at the vertex, and vertex is
And
And that's the required dimensions .
AB^2 + BC^2 = AC^2
AB^2 + 6^2 = square root 117^2
AB^2 + 36 = 117
Now subtract 117 from both sides
AB^2 = 81
AB = square root 81 = 9
Therefore AB is 9 cms.
AB^2 + 6^2 = square root 117^2
AB^2 + 36 = 117
Now subtract 117 from both sides
AB^2 = 81
AB = square root 81 = 9
Therefore AB is 9 cms.
Answer:
Step-by-step explanation:
Given equation can be re written as
............(i)
Now it is given that y(π/2) = 2e
Applying value in (i) we get
ln(2e) = sin(π/2) + c
=> ln(2) + ln(e) = 1+c
=> ln(2) + 1 = 1 + c
=> c = ln(2)
Thus equation (i) becomes
ln(y) = sin(x) + ln(2)
ln(y) - ln(2) = sin(x)
ln(y/2) = sin(x)