Peter has 3200 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed a
2 answers:
Answer:
Step-by-step explanation:
Expression for the rectangular area and perimeter are, respectively:
After some algebraic manipulation, area expression can be reduce to an one-variable form:
The first derivative of the previous equation is:
Let the expression be equalized to zero:
The second derivative is:
According to the Second Derivative Test, the critical value found in previous steps is a maximum. Then:
The maximum area is:
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Answer:
(a) The probability density function of <em>X</em> is:
(b) The value of P (129 ≤ X ≤ 146) is 0.3462.
(c) The probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.
Step-by-step explanation:
The random variable <em>X</em> is defined as the flight time between the two cities.
Since the random variable <em>X</em> denotes time interval, the random variable <em>X</em> is continuous.
(a)
The random variable <em>X</em> is Uniformly distributed with parameters <em>a</em> = 10 minutes and <em>b</em> = 154 minutes.
The probability density function of <em>X</em> is:
(b)
Compute the value of P (129 ≤ X ≤ 146) as follows:
Apply continuity correction:
P (129 ≤ X ≤ 146) = P (129 - 0.50 < X < 146 + 0.50)
= P (128.50 < X < 146.50)
Thus, the value of P (129 ≤ X ≤ 146) is 0.3462.
(c)
It is provided that a randomly selected flight between the two cities will be at least 3 minutes late, i.e. <em>X</em> ≥ 128 + 3 = 131.
Compute the value of P (X ≥ 131) as follows:
Apply continuity correction:
P (X ≥ 131) = P (X > 131 + 0.50)
= P (X > 131.50)
Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.