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A common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, becau

Serga [27]

Answer:

0.0159

Step-by-step explanation:

Given that a common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, because customers who buy tickets do not always show up for the flight.

Here if X is the no of persons that do not show up, then X is binomial as each trial is independent with p = 0.04 and n =150 (no of tickets sold)

The plane is overbooked if more than 150 show up

i.e. less than 2 do not show up

Hence the probability that the airline overbooked this flight

= $P(X$

4 0

3 years ago

1 + tanx / 1 + cotx =2

Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer: x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0

3 years ago

The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?

egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola. The first few steps are as follows. Set the parabola equal to 0 so you can solve for the vertex. Separate the x terms from the constant by moving the constant to the other side of the equals sign. The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out). Let's start there. The first 2 steps result in this polynomial:

$-2x^2+4x=-1$ . Now we factor out the -2:

$-2(x^2-2x)=-1$ . Now we complete the square. This process is to take half the linear term, square it, and add it to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. We add 1 into the set of parenthesis. But we actually added into the parenthesis is +1(-2). The -2 out front is a multiplier and we cannot ignore it. Adding in to both sides looks like this:

$-2(x^2-2x+1)=-1-2$ . Simplifying gives us this:

$-2(x^2-2x+1)=-3$

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex. Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

$-2(x-1)^2+3=y$