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3 years ago

Which of the following equations has an infinite number of solutions?

Mathematics

1 answer:

Inessa [10]3 years ago

5 0

It is 3x +5 = 2x -1 + x
i hope that is was some sort of help you are welcome

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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

The diameter of a circular garden is 22 feet. What is the approximate area of the shaded region

SOVA2 [1]

Answer:

Step-by-step explanation:

The radius it .5 of the diameter so in this case the radius is 11. The formula for area is pi*r^2= pi*11^2=pi*121=380.13

7 0

3 years ago

Read 2 more answers

If 5/8 yards of fabric costs 3.75 dollars what does one yard cost

Sladkaya [172]

3.75 / 5 =0.75
0.75 x 8 = 6
So one yard is 6
Sorry, I’m not American so I don’t know about dollars but this should be correct x

4 0

3 years ago

Amanda is a cabinetmaker. She makes cabinets using one of three types of wood: birch, maple, or cherry. She can finish each type

padilas [110]

Assume that each material in each category has an equal chance of being selected.

Categories:
(a) Type of wood used to make the cabinet: birch, maple, cherry.
Each type of wood has a probability of 1/3 to be selected.
(b) Type of finish: transparent, semi-transparent.
Each type of finish has a probability of 1/2 to be selected.
(c) Knob: bronze, steel, wood.
Each material has 1/3 probability to be selected.

Each event (selecting material for the cabinet, for the finish, and for the knob) is an independent event.

Therefore
Probablity(birch wood AND bronze knob) = (1/3)*(1/3) = 1/9
Probability(wood knob) = 1/3
Probability(transparent stain) = 1/2
Probability(cheryy wood AND semi-transparent stain) = (1/3)*(1/2) = 1/6

6 0

3 years ago

What is the solution to this system of equations? 3x + 2y=45 , 4x-y=5

sleet_krkn [62]

The solution is (5, 15)

Multiply the second equation by 2 and then add through.

3x + 2y = 45
8x - 2y = 10
-----------------
11x = 55
x = 5

Then plug in to get the y value

3(5) + 2y = 45
15 + 2y = 45
2y = 30
y = 15

3 0

3 years ago