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Zigmanuir [339]
3 years ago
5

A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in

36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?A. 3B. 4C. 6D. 9E. 12
Mathematics
1 answer:
kozerog [31]3 years ago
8 0

Answer: C.  6

Step-by-step explanation:

Given : Time taken by Machine R to do the job = 36 hours

Time taken by Machine S to do the job = 18 hours

Let x be the number of each type of machine to do the job.

Then , according to the question the required equation will be :-

\dfrac{x}{36}+\dfrac{x}{18}=\dfrac{1}{2}\\\\\Rightarrow\ x(\dfrac{1}{36}+\dfrac{1}{18})=\dfrac{1}{2}\\\\\Rightarrow\ \dfrac{x}{12}=\dfrac{1}{2}\\\\\Rightarrow\ x=\dfrac{12}{2}=6

Hence, the number of machines of type R were used =6

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Tay–Sachs Disease Tay–Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri
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Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used 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 probability that their offspring will develop the disease is approximately .25.

This means that p = 0.25

Three children:

This means that n = 3

Question a:

This is P(X = 3). So

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

P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

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

P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tay–Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

p = 0.75*0.75*0.25 = 0.1406

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

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When her number is divided by 12, the remainder is 6, what is the remainder when the number is idvided by 9?
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Remainder: 6 when divided by 12
So 12x6= 72
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Solve for the roots in the following equation. hint: factor both quadratic expressions. (x 4 5x 2 - 36)(2x 2 9x - 5) = 0
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Hello :
 <span>(x 4 +5x²  - 36)(2x ²+ 9x - 5) = 0</span>

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Answer:

5 chickens

Step-by-step explanation:

Let's set chickens and pigs with variables "c" and "p" respectively.

We know there a 16 animals in all, so there are:

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Now we have our system of equations:

c+p=16

2c+4p=54

We are trying to find the number of chickens, so we write the first equation in respect to c.

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Substituting the first derived equation to equation two:

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Simplifying:

64-2c=54

2c=10

c=5

So there are 5 chickens.

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