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timama [110]
3 years ago
10

If y varies directly as x and y=60 when x=12, find y if x=35

Mathematics
2 answers:
natita [175]3 years ago
5 0

Answer:

y = 175

Step-by-step explanation:

Case condition: y ∝ x

∴ y = 60  when x = 12

  y = ?     when x = 35

since the relationship is direct, then use cross multiplication method.

60 = 12

?  = 35

Therefore,

? x 12 = 35 x 60

? x 12 = 2100

divide through by 12 to make '?' the subject

? = 2100/12

? = 175

Therefore Y = 175 when x = 35

Drupady [299]3 years ago
3 0

Answer: y=175

Step-by-step explanation: y&x

y=kx

(Where k is a constant)

When y=60,x=12...we have

60=12k

Divide both side by 12

12k/12=60/12

k=5

Equation becomes y=5k

Finding y when x=35

y=5(35)

y=175

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Answer:

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

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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:

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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

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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?

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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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