The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 m
1 answer:
Answer:
0.35
Step-by-step explanation:
we know that
The formula to use for uniform distribution is equal to
where,
b is the upper limit of the distribution which is 5 minutes in this case.
a is the lower limit of the distribution which is 0 minutes in this case.
x is the concerned value which is 1.75 minutes in this case.
substitute the given values
That means
The probability that a randomly selected passenger has a waiting time less than 1.75 minutes is 0.35
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SOLUTION
From the question, we are told that y varies inversely as x
This is mathematically written as
Now, we will remove the proportionality sign and replace it with equal to sign =
If we do this, we will intoduce a constant k
So we have the formula
We will substitute the values of x for 10 and y for 8 into the formula to get k, we have
Now, we will substitute k for 80 back into the formula to get the inverse function, we have
Hence the answer is option C
Given:
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter
Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l (1)
Part A.
The area is
A = w*l
= (2 - l)*l
A = 2l - l²
This is a quadratic function or a parabola.
Part B.
Write the parabola in standard form.
A = -[l² - 2l]
= -[ (l -1)² - 1]
= -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.
The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).
Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter
Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l (1)
Part A.
The area is
A = w*l
= (2 - l)*l
A = 2l - l²
This is a quadratic function or a parabola.
Part B.
Write the parabola in standard form.
A = -[l² - 2l]
= -[ (l -1)² - 1]
= -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.
The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).
Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.