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3 years ago

What is the distance between -4 and 7 on a number line?

Mathematics

2 answers:

tiny-mole [99]3 years ago

7 0

Answer:

Step-by-step explanation:

Lina20 [59]3 years ago

6 0

Answer:

Step-by-step explanation:

Take -4 and turn it to 0.

-4 + 4 = 0

We added 4 to the number

0 + 7 = 7

We added 7 to the number.

4 + 7 = 11

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

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

If I equals PRT which equation is equivalent to T

snow_lady [41]

I = P · R · T
Divide both sides by P·R
I/(P·R) = T

Answer: $\frac{I}{PR} = T$

7 0

3 years ago

At town has a population of 13000 and grows at 2.5% every year. What will be the population after 5 years, to the nearest whole

zubka84 [21]

Answer:

the answer will be 14708

4 0

2 years ago

Read 2 more answers

Ralph is 3 times as old as Sarah. in six years, Ralph will be only twice as old is Sarah will be then. find Ralphs age now.

Phantasy [73]

First, we start with the equation that the problem told us, which is:

R + 6 = 2 * (x + 6)

Then, we distribute the two:

R + 6 = 2x + 12

Now, we put R on its own side.

R = 2x +6

So, the answer must be C, 2x + 6

8 0

3 years ago

Read 2 more answers

A ramp into a building forms a 6° angle with the ground. If the ramp is 8 feet long, how far away from the building is the entry

Sauron [17]

Answer:

7.96 ft

Step-by-step explanation:

Given;

Length of ramp L = 8 ft

Angle with the horizontal (ground) = 6°

Applying trigonometry;

With the length of ramp as the hypothenuse,

The horizontal distance d as the adjacent to angle 6°

Since we want to calculate the adjacent and we have the hypothenuse and the angle. We can apply cosine;

Cosθ = adjacent/hypothenuse

Substituting the values;

Cos6° = d/8

d = 8cos6°

d = 7.956175162946

d = 7.96 ft

The building is 7.96ft away from the entry point of the ramp.

6 0

3 years ago