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Calculus and probability

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The graph of a function f: [-2, p] -> R is shown below.
The average value of f over the interval [-2, p] is zero.
The area of the shaded region is 25/8.
If the graph is a straight line, for 0 <= x <= p, then the value of p is:

a. 2
b. 5
c. 5/4
d. 5/2
e. 25/4

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is:
a. 8
b. 9
c. 10
d. 11
e. 12

The sample space when a fair die is rolled is {1, 2, 3, 4, 5, 6}, with each outcome being equally likely. For which of the following pairs of events are the events independent?
a. {1. 2. 3} and {1, 2}
b. {1, 2} and {3, 4}
c. {1, 3, 5} and {1, 4, 6}
d. {1, 2} and {1, 3, 4, 6}
e. {1, 2} and {2, 4, 6}

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https://brainmass.com/math/probability/calculus-and-probability-635771

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1. The average value of a function over a finite interval is given by the integral
of that function over that interval divided by the length of the interval.

The integral of f over the interval [-2, p] is equal to the sum of the integral of f
over the interval [-2, 0] and the integral of f over the interval [0, p].

The area of the shaded region is given as 25/8. Notice that that area is equal
to the integral of the function 0 - f (= -f) over the interval [-2, 0].
Therefore, the integral of f over the interval [-2, 0] is equal to -25/8.

Since the integral of f over the interval [-2, p] is 0, the integral of f
over the interval [0, p] is equal to 0 - (-25/8) = 25/8.

Suppose the graph of f on the interval [0, p] is a straight line.
The equation of a straight line is y = mx + b, where m is the slope
and b is the y-coordinate of the y-intercept.

Since the graph passes through the origin (i.e., through the point (0, 0)),
the y-intercept is the origin, and its y-coordinate is 0. Therefore, b = 0, so
the equation of the line is y = mx where m is its slope.

The graph passes through the point (p, p),
so substituting p for both x and y in the equation y = mx, we get

p = mp

Dividing both sides of this equation by p, we obtain

1 = m

Thus the equation of the line is y = x. Therefore, f(x) = x on the interval [0, p].
Now an antiderivative of the function f(x) is (x^2)/2,
so the integral of f over the interval [0, p] is [(p^2)/2] - [(0^2)/2],
which is just (p^2)/2 since (0^2)/2 = 0.

We know that the value of the integral of f over the interval [0, p] is 25/8,
so to find the value of p, we need to equate (p^2)/2 to 25/8 and solve for p:

(p^2)/2 = 25/8

Multiplying both sides by 2:

p^2 = 25/4

Therefore, |p| = sqrt{25/4} = 5/2.

We know that p is positive (because the point (p, p) lies to the right of the origin), so p = 5/2.

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3. Since f'(x) is an antiderivative of ...

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