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shtirl [24]
3 years ago
9

Customers at Beans Coffee Shop have entered their name in a drawing for one of 4 gift cards. The gift cards each have a value of

$10. If 79 customers enter the drawing, what is the total number of ways Beans Coffee Shop could choose the winners?
Mathematics
1 answer:
Dimas [21]3 years ago
8 0

Answer:

36,060,024 ways

Step-by-step explanation:

To choose the winner in this case, the order is very important because it determines who is chosen first  out of the seventy-nine customers whose names were entered for the draw.

The above supports the reason Permutation will be used instead of Combination.

The total number of ways is given by:

⁷⁹P₄ = 79!/(79-4)!

       =  79!/75!

       = [79 x 78 x 77 x 76 x75!]/75!

       = 79 x 78 x 77 x 76

       =   36,060,024 ways

This implies that there are  36,060,024 ways the Bean Coffee Shop could choose the winners.

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Need Help! 10 points!
larisa86 [58]

The answer is 1680. To get this you do 8*7*6*5 because there are 8 options for the first character, and then you can repeat that for the second one with 7, 6, and 5

6 0
3 years ago
A softball team is ordering pizza to eat after their tournament. They plan to order cheese pizzas that cost $6 each and four-top
Softa [21]

Answer:

46

Step-by-step explanation:

Well $1o each than 10×4=40 and 40 +6=46

7 0
3 years ago
Read 2 more answers
If f(x) = |x| + 9 and g(x) = –6, which describes the range of (f + g)(x)?
Leokris [45]
Just want points byeeee hope u fail jkjkjk
3 0
3 years ago
if there are four trucks parked in line and after the first truck each truck weighed 2 more tons than the one before. They all w
algol13
Each truck weighs five tons. 

1st truck: n 
2nd truck: n + 2
3rd truck: n + 4
4th truck: n + 6 

n + n + 2 + n + 4 + n + 6 = 32 
4n + 12 = 32 
      - 12   - 12
--------------------
4n = 20 
----  ------
 4      4

n = 5 
6 0
3 years ago
Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variable
IgorLugansk [536]

Answer:

P(T_A < T_B) = P(T_A -T_B

Step-by-step explanation:

Assuming this problem: "Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are independent. Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs".

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we can create some notation.

Let A the values for Ann we know that n1 = 30 jobs solved in sequence and we can assume that the random variable X_i the time in order to do the ith job for i=1,2,....,n_1. We will have the following parameters for A.

\mu_A = 50, \sigma_A =10

W can assume that B represent Bob we know that n2 = 30 jobs solved in sequence and we can assume that the random variable [tex[X_i[/tex] the time in order to do the ith job for i=1,2,....,n_2. We will have the following parameters for A

\mu_B = 52, \sigma_B =15

And we can find the distribution for the total, if we remember the definition of mean we have:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

And T =n \bar X

And the E(T) = n \mu

Var(T) = n^2 \frac{\sigma^2}{n}=n\sigma^2

So then we have:

E(T_A)=30*50 =1500 , Var(T_A) = 30*10^2 =3000

E(T_B)=30*52 =1560 , Var(T_B) = 30 *15^2 =6750

Since we want this probability "Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs" we can express like this:

P(T_A < T_B) = P(T_A -T_B

Since we have independence (condition given by the problem) we can find the parameters for the random variable T_A -T_B

E[T_A -T_B] = E(T_A) -E(T_B)=1500-1560=-60

Var[T_A -T_B]= Var(T_A)+Var(T_B) =3000+6750=9750

And now we can find the probability like this:

P(T_A < T_B) = P(T_A -T_B

P(\frac{(T_A -T_B)-(-60)}{\sqrt{9750}}< \frac{60}{\sqrt{9750}})

P(Z

7 0
3 years ago
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