Ples help and ples give derect awnser not a link you will get brainly if u do

A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome

Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

$=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135$

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

$=P(X$

Thus, the probability that the flight has empty seats is 0.4625.

4 0

3 years ago

JKL is a straight line.

kirza4 [7]

Answer:

x = 32°

Step-by-step explanation:

∆KLM is an isosceles triangle because it has two equal sides, KL & KM. Therefore, the angles opposite to each of the two equal sides, which are referred to as the base angles are congruent to each other.

m<KML = m<KLM = 58°

m<MKL = 180 - (58 + 58) (Sum of triangle)

m<MKL = 64°

m<JKM = 180 - m<MKL (linear pair theorem)

m<JKM = 180 - 64 (Substitution)

m<JKM = 116°

∆JKM is also an isosceles triangle with two equal sides. Therefore, it's based angles (x & <J) would also be equal to each other.

Thus:

x = ½(180 - m<JKM)

x = ½(180 - 116) (Substitution)

x = 32°

3 0

2 years ago

In which quadrant does the point with c ordinate (4, -3) lie?

KiRa [710]

Answer:

4

Step-by-step explanation:

hope the diagram helps haha

3 0

2 years ago

Last question of my homework pls <3

dimaraw [331]

Answer:

J(2, 5 )

Step-by-step explanation:

Using the section formula

$x_{J}$ = $\frac{1(-2)+2(4)}{2+1}$ = $\frac{-2+8}{3}$ = $\frac{6}{3}$ = 2

$y_{J}$ = $\frac{1(1)+2(7)}{2+1}$ = $\frac{1+14}{3}$ = $\frac{15}{3}$ = 5

Thus J = (2, 5 )

7 0

2 years ago

Simplify the expression. tan(sin^−1 x)

Blizzard [7]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2799412

_______________

Let $\mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}$

(that is the range of the inverse sine function).

So,

$\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}$

Square both sides:

$\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}$

Since $\mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},}$ then $\mathsf{cos\,\theta}$ is positive. So take the positive square root and you get

$\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}$

Then,

$\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}$

I hope this helps. =)

Tags: inverse trigonometric function sin tan arcsin trigonometry

3 0

3 years ago

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