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sveta [45]
3 years ago
11

Line segment AB has a length of 4 units. It is translated 2 units to the right on a coordinate plane to obtain line segment A'B'

. What is the length of A'B'?
Mathematics
2 answers:
tatyana61 [14]3 years ago
8 0

Question

Line segment AB has a length of 4 units. It is translated 2 units to the right on a coordinate plane to obtain line segment A'B'. What is the length of A'B'?

Answer:

The Answer is 4 Units!

Step-by-step explanation:

length of the line segment does not change. you are just moving it 2 units to the right.

Georgia [21]3 years ago
4 0

Answer:

length of the line segment does not change. you are just moving it 2 units to the right. answer would be 4 units

Step-by-step explanation:

Hope this helps!

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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

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