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3 years ago

6

The biology faculty at a college consists of 9 full professors, 14 associate professors, 13 assistant professors, and 4 instruct

ors. if one faculty member is randomly selected, find the probability of choosing a full professor or an instructor

1 answer:

-Dominant- [34]3 years ago

5 0

32 total
9/32 -professor
9/32 -instructor
9/32+9/32=18/32=9/16

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3 years ago

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Find the median of the following set of data 12,20,1,20,8

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Answer:

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Step-by-step explanation:

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3 years ago

Substitution for:<br> X-5y=10<br> 2x-10y=20

sergey [27]

Since you need an isolated variable to use the substitution method, we need to re-arrange one of the equations. This will probably be easiest to do with the first one.
Add 5y to both sides of the first equation.
x=10+5y
Now, in the second equation, put in 10+5y in any spot that has an x.
2(10+5y)-10y=20
Distribute the 2 to both numbers in the parenthesis.
20+10y-10y=20
Combine like terms.
20=20
This means that the two equations are actually the same. You can see this if you multiply the whole first equation by 2
2(x-5y=10)
2x-10y=20, which is the same as the second equation. Therefore, the two equations are actually the same one.

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3 years ago

A certain firm has plants A, B, and C producing respectively 35%, 15%, and 50% of the total output. The probabilities of a non-d

Sliva [168]

Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

$P(A) = P_{1} + P_{2} + P_{3}$

In which $P_{1}$ is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

$P_{1} = 0.35*0.25 = 0.0875$

$P_{2}$ is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

$P_{2} = 0.15*0.05 = 0.0075$

$P_{3}$ is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

$P_{3} = 0.50*0.15 = 0.075$

So

$P(A) = 0.0875 + 0.0075 + 0.075 = 0.17$

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

$P = \frac{0.5*0.15}{0.17} = 0.4412$

There is a 44.12% probability that the defective product came from C.

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3 years ago

A student reads 3/5 of a book in 30 minutes how much of the book will be read if the student reads at the same speed for 40 minu

erastova [34]

Answer:

4/5 of the book in 40 mins

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2 years ago