Which value is equivalent to the expression 2^3 + 3^4

Occasionally an airline will lose a bag. Suppose a small airline has found it can reasonably model the number of bags lost each

julia-pushkina [17]

Answer:

a) The probability that the airline will lose no bags next monday is 0.1108

b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227

c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2

Step-by-step explanation:

The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula

$P(X=k) = \frac{e^{-2.2} * {2.2}^k }{k!}$

a)

$P(X=0) = \frac{e^{-2.2} * {2.2}^0 }{0!} = 0.1108$

The probability that the airline will lose no bags next monday is 0.1108.

b) Note that $P(X \in \{0,1,2\} = P(X=0) + P(X=1) + P(X=2)$ . And

$P(X=0)+P(X=1)+P(X=2) = e^{-2.2} * (1 + 2.2 + 2.2^2/2) = 0.6227$

Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.

c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.

6 0

3 years ago

An object on the moon weighs about 0.17 of what it would weigh on earth.

Ivahew [28]

The answer would be E.

All you have to do is multiply 185 by 0.17 and you get your answer, which would be E.

3 0

3 years ago

Read 2 more answers

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$