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IgorLugansk [536]
3 years ago
6

You roll a single die numbered from 1 to 6 twice. The probability of rolling an even number the first time and a 6 the second is

the setup. What is the probability "rolling an even number the first time"? (Answer in percentage)
Mathematics
1 answer:
Arlecino [84]3 years ago
7 0

Answer:

<h2>The answer is 50%</h2>

Step-by-step explanation:

step one :

first and foremost we are going to list the sample space related to a die

the same space is  S= {1,2,3,4,5,6}= 6

step two:

now the even numbers in the sample space are ={2,4,6}= 3

hence we can solve for the probability of obtaining an even number to be

Pr(even number)= 3/6= 1/2.

step three:

since we are to provide the answer in percentage format, we can easily do this by multiplying the answer by 100 to get the percentage

(1/2)*100= 0.5*100= 50%

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A local club is arranging a charter flight to Hawaii. The cost of the trip is ​$586 each for 80 ​passengers, with a refund of​ $
professor190 [17]

Answer:

a) The number of passengers that will maximize the revenue received from the flight is 99.

b) The maximum revenue is $48,609.

Step-by-step explanation:

We have to analyse two cases to build a piecewise function.

If there are 80 or less passengers, we have that:

The cost of the trip is $586 for each passenger. So

R(n) = 586n

If there are more than 80 passengers.

There is a refund of $5 per passenger for each passenger in excess of 80. So the cost for each passenger is

R(n) = (586 - 5(n-80))n = -5n^{2} +400n + 586n = -5n^{2} + 986n.

So we have the following piecewise function:

R(n) = \left \{ {{586n}, n\leq 80 \atop {-5n^{2} + 986n}, n > 80} \right

The maxium value of a quadratic function in the format of y(n) = an^{2} + bn + c happens at:

n_{v} = -\frac{b}{2a}

The maximum value is:

y(n_{v})

So:

(a) Find the number of passengers that will maximize the revenue received from the flight.

We have to see if n_{v} is higher than 80.

We have that, for n > 80, R(n) = -5n^{2} + 986n, so a = -5, b = 986

The number of passengers that will maximize the revenue received from the flight is:

n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6

Rounding up, the number of passengers that will maximize the revenue received from the flight is 99.

(b) Find the maximum revenue.

This is R(99).

R(n) = -5n^{2} + 986n

R(99) = -5*(99)^{2} + 986*(99) = 48609

The maximum revenue is $48,609.

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