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3 years ago

90 c + 62/9 , when c = 37

1 answer:

8 0

So evalueate
90c+62/9
90c means 90 times c
if c=37 then
subsiutte
90 times 37+62/9
3330+62/9

62/9=6 and 8/9

3330+6 and 8/9
6 and 8/9=6+8/9
3330+6+8/9
3336 and 8/9

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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

Multiply ( + 2) (2 – )

Elina [12.6K]

Answer:

-4

Step-by-step explanation:

('+')('-') = ('-')

Therefore,

(+2)(-2) = (-4)

7 0

2 years ago

I NEED HELP IM IN CLASS RN PLEASEE HELP! WILL GIVE BRIANLIEST!

fgiga [73]

Answer:

y=x+12

Step-by-step explanation:

Well the slope is -4/-4 =1

We plug it into the formula, y=mx+b, but m=1

So it is y=1x+b

We know the points, so we plug y and x in which is:

-4=-4*4+b

-4=-16+b

b=12

So the point slope form of that is y=x+12

4 0

3 years ago

The edges of each tissue box measure 6.5 inches the dimensions of the shipping container are 19.5 inches by 39 inches by 19.5 in

alexdok [17]

The number of tissues that can fit the container will be given by:
[volume of the container]/[volume of tissue box]
volume of container:
V=length×width×height
V=19.5×39×19.5
V=14829.75

Volume of tissue box:
v=6.5×6.5×6.5
v=274.625

Number of tissues in the container:
14829.75/274.625
=54

Answer: 54 tissue paper boxes

3 0

3 years ago

Show how to simplify this problem using the distributive property. 4(12 + 6)

Sunny_sXe [5.5K]

Hey there!

4(12 + 6)

4(12) + 4(6)

4(12) = 48

4(6) = 24

48 + 24 = ?

48 + 24 = 72

Answer: 72 ☑️

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

5 0

3 years ago

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