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Ne4ueva [31]
3 years ago
7

Jerry is saving money for a job. After the first 3 weeks he saved $135. Assuming the situation is proportional. Use the unit Rae

to write an equation rearing the amount saved s to the number of weeks w worked. At this rate how much will Jeremiah have after 8 weeks?
Mathematics
1 answer:
VMariaS [17]3 years ago
3 0

The equation would be s = 45w and she would have 360 in 8 weeks.

In order to find this, we first have to find the per week amount. To find this, we divide the amount by the number of weeks.

135/3 = 45

Now that we have this we can multiply that by the variable and set equal to the total amount.

s = 45w

Now to find the amount in 8 weeks, simply plug in the 8 to w.

s = 45w

s = 45(8)

s = 360

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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: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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

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

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

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}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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}

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

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}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

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}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
Solve pls brainlies
MrRa [10]

Answer:

2) 11

3) 21

4) 7

Step-by-step explanation:

2) (-6) -(-17) = 17 - 6 = 11

3) 19 - (-2) = 19 + 2 = 21

4) (-13) + 20 = 20 - 13 = 7

6 0
2 years ago
What is the answer to this question?<br> I am confused.
olga nikolaevna [1]

Answer:

y=88 x=42

Step-by-step explanation:

Since the total measurments of one triangle have to equal 180 and one angle is already 50, and the angle on the oppisite side of the y angle is 90 the only logical number for y would be 88. Hope this helped :)

5 0
3 years ago
A turtle and a snail are 300 feet apart when they start moving towards each other. The turtle is moving at a speed of 5 feet per
givi [52]

Answer:

It Is 60

Step-by-step explanation:

300/5=60x1=60

3 0
3 years ago
Read 2 more answers
(b) The area of a rectangular pool is 5488 m\2 If the length of the pool is 98 m, what is its width?​
mrs_skeptik [129]

Answer:

56

Step-by-step explanation:

To find the area of a rectangle we have the foruma A=WxL.

But we already have the area and length so we can plug that in

5488=Wx98

Now its an algebreic expression.

SInce its multiplying we do the opposite, so we divide 98 on both sides.

98/98 crosses itself out so now its 5488/98. Which equals 56. So now our expression is W=56. To fact check we put the numbers 56 and 98 into the formula to see if we get 5488.

A=56x98

A=5488

7 0
1 year ago
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