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skad [1K]
3 years ago
8

A builder could get 6 sheets of sheetrock for 3$. If he bought 60 sheets , how much money would he have spent

Mathematics
1 answer:
Andrei [34K]3 years ago
7 0
The builder would have spent 20$.
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Jerry was comparing the points the bulls socered for different games he recorded 85,85, 86,77, 93 determine the mean median mode
maks197457 [2]
Answers:
Mean - 85.2
Median - 85
Mode - 85

Explanation:
Mean - 77+85+85+86+93=426/5 (the amount of data points given) = 85.2

Median - Organize the data points from least to greatest and eliminate the outermost numbers two at a time (one from both side) until you’re left with one.

Mode - The number that appears the most often in a given data set
3 0
3 years ago
Evaluate 2A - 3B for A = 6 and B = -3.
Rasek [7]
2 (6) - 3 (-3)  , 12 - -9 , 12 + 9 , 21
6 0
3 years ago
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Accuracy in taking orders at a drive-through window is important for fast-food chains. Periodically, QSR Magazine publishes "The
pav-90 [236]

Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that p = 0.905

You and 2 friends:

So 3 people in total, which means that n = 3

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

P(X \geq 1) = 1 - P(X = 0)

With what we found in b:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

p = 0.905*0.095*0.095 = 0.0082

0.0082 = 0.82% probability that only your order will be filled correctly

7 0
3 years ago
The graph compares the weights in pounds of 100 dogs and cats that are brought in to a veterinarian's office.
GenaCL600 [577]
In case of dogs the value 10 is the minimum value. So all the values lie above 10. In total there were 100 dogs.
So for dogs, we can say number of dogs above the value of 10 pound are 100.

In case of Cats, 10 lies at the position of median. Median is the central value and 50% values lie above the median value. So number of cats with weight above 10 pound is 50.

Thus, we can conclude that there were 50 more dogs than the cats with weight over 10 pounds. So option C gives the correct answer.
8 0
3 years ago
Read 2 more answers
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mestny [16]
Hey :) it’s (2,5) it’s C !!
6 0
3 years ago
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