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2 years ago

Jamil pays $20.00 for dinner. He decides to leave a 15% tip on top. How much should he pay in total

Mathematics

2 answers:

Helga [31]2 years ago

6 0

Answer:

$23.00

Step-by-step explanation:

The tip amount is $3.00, but the total is $23.00

Juli2301 [7.4K]2 years ago

3 0

Answer:

he should pay $60

Step-by-step explanation:

$20.00 + $40.00 = $60.00

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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 value of a $3000 computer decreases about 30% each year. write a function for the computers value V(t)

den301095 [7]

Answer:

Function for given situation is : $V(t)=3000(0.70)^t$

Value of computer after 4 years = $720.3.

Step-by-step explanation:

Given that the value of a $3000 computer decreases about 30% each year. Now we need to write a function for the computers value V(t). then we need to find about how much will the computer be worth in 4 years.

It clearly says that value decreases so that means function represents decay.

For decay we use formula:

$A=P(1-r)^t$

where P=initial value = $3000,

r= rate of decrease =30% = 0.30

t= number of years

A=V(t) = future value

so the required function is $V(t)=3000(1-0.30)^t$

or $V(t)=3000(0.70)^t$

Now plug t=4 years to get the value of computer after 4 years.

$V(4)=3000(0.70)^4$

$V(4)=720.3$

Hence final answer is $720.3.

7 0

2 years ago

Read 2 more answers

An ice cream truck began its daily route with 95 gallons of ice cream. The truck driver sold 78% of the

Brilliant_brown [7]

Answer:

74.1 or 74

Step-by-step explanation:

3 0

2 years ago

The numbers x and y are inversely proportional. When the sum of x and y is 42, x is twice y. What is the value of y when x = -8?

dolphi86 [110]

Answer:

y= -4

Step-by-step explanation:

x+y=42. We know that x plus y equals 42.

x=2y . We also know that x equals twice y.

2y+y=42. Since we know that x=2y, we're going to replace x with 2y from the original equation, simplified would be 3y=42.

y=14. 3y=42 simplified is y=14, this is because we divided both sides by 3 (since 3 and 42 are both divisible by 3) (btw divisible by 3 simply means that it can be divided by 3)

x+14=42. Now that we know what y equals, let's replace y with 14 from the original equation.

x=28. Now that we know that x=28 and y=14, we know that y is always x/2.

When x=-8, we divide it by 2 and get, -4.

Not absolutely positive as this was all mental math, but hope this helps! :)

4 0

3 years ago

The endpoints of GH are G(10,1) and H(3,5). What is the midpoint of GH?

svet-max [94.6K]

Answer:

The answer is (6.5,3)

Step-by-step explanation:

The solution is in the image

8 0

1 year ago