4x^-2*y/x*y^-2 step by step

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).

$P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918$

3 0

3 years ago

M is the centroid of the triangle.<br> CM=7, PM=10, & BQ=18<br> M<br> AM=[?]<br> Is it 6?

Snowcat [4.5K]

Answer:

5

Step-by-step explanation:

6 0

3 years ago

Mass of a Robin was 72.89 grams the mass of a blue Jay was 84.28 grams what is the estimate difference in the mass of the birds

pentagon [3]

Answer:

Step-by-step explanation:

estimation: 84-73

11

5 0

3 years ago

The sum of 5 consecutive odd numbers is 505

pentagon [3]

Answer:

In a sequence of 5 numbers, the 3rd number is the middle number. So in your problem the third number is the average of the 5 numbers -- which is the sum of the numbers, divided by how many there are: 505/5 = 101.

Explanation:

8 0

3 years ago

In △ABC , line AB is extended up to point E and line BC is extended up to D . The measure of some of the angles is given as; ∠BA

Over [174]

Answer:

$\angle ACD = 130 ^\circ$ .

Step-by-step explanation:

$\angle ABC$ and $\angle EBC$ are supplementary angles. Their sum should be $180^\circ$ . $\angle EBC = 120^\circ$ $\implies \angle ABC = 180^\circ - \angle EBC = 180^\circ - 120^\circ = 60^\circ$ .

The sum of the three angles in triangle $\triangle ABC$ : $(\angle ABC + \angle BAC + \angle ACB)$ should be equal to $180^\circ$ .

$\angle ABC = 60^\circ$ and $\angle BAC = 70^\circ$ $\implies \angle ACB = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 70^\circ - 60^\circ = 50^\circ$ .

$\angle ACB$ and $\angle ACD$ form another pair of supplementary angles. Their sum should also be $180^\circ$ . $\angle ACB = 50^\circ$ $\implies \angle ACD = 180^\circ - \angle ACB = 180^\circ - 50^\circ = 130^\circ$ .

8 0

4 years ago