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MatroZZZ [7]
3 years ago
10

An animal is randomly selected from a group consisting of 5 cats, 11 dogs, and 14 rabbits. Find the probability of the outcome.

Enter your answer as a decimal, rounded to the nearest hundredth.
At a soccer camp, a group of soccer players randomly drew numbers from 1 to 44 to determine which practice squad they would be on. What is the probability that the first player to chose a number picked a number less than 12? write answer as a decimal rounded to the nearest hundredth.




***Probability*** please help fast
Mathematics
1 answer:
ohaa [14]3 years ago
5 0
1. There are 30 animals, so the probability of picking a cat is 5/30, or 1/6. (o.16)
The probability of picking a dog is 11/30 or (0.36)
The probability of picking a rabbit is 14/30 or 7/15 (0.46)

2. The probability of picking a number less than 12 is 11/44, or 1/4 (0.25)
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The Continuing Education Division at the Ozark Community College offers a total of 30 courses each semester. The courses offered
dybincka [34]

Answer:

The college should offer 20 practical courses and 10 humanistic courses in order to maximize the revenue.

Step-by-step explanation:

a)

Let x and y be  

<em>x= number of practical courses the college will offer</em>

<em>y= number of humanistic courses the college will offer</em>

we have the following inequalities

x ≥ 10

y ≥ 10

x+y = 30

The only points (x,y) that satisfy all this inequalities are (10,20) and (20,10) (see picture attached).

On the other hand, the revenues would be given by  

R = 1,500x + 1,000y

The maximum of R is attained in (x,y) = (20,10) and is  

R = 300,000 + 100,000 = 400,000

and the college should be offering 20 practical courses and 10 humanistic ones.

b)  

If x = 21 then y must be 9, and then y does not satisfy y ≥ 10.

So you can only offer additional humanistic courses.

In n is an integer < 30

But if y=10+n  then x must be 30-n and the revenue would be

R = 1,500(30-n) + 1,000(10+n) =  

350,000 - 1,500n + 100,000+ 1,000n =  

400,000 - 500n < 400,000

This result means in terms of offering additional courses, that is not worth doing it.

8 0
3 years ago
Please Explain how you got this answer
liberstina [14]
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</span>When the order doesn't matter, it is a Combination<span>.
</span>
The formula means <span>the number of ways to combine k items from a set of n

***************************************

n = 4 (brisket, chicken, ribs, turkey)
k = 2 (two-meat combos)

</span>_nC_k= \frac{n!}{k!(n-k)!} &#10; \\  \\ _nC_k= \frac{4!}{2!(4-2)!}&#10;\\ \\ _nC_k= \frac{4!}{2!2!}&#10;\\ \\_nC_k= \frac{4*3*2*1}{2*1*2*1}&#10;\\\\ _nC_k= \frac{4*3}{2*1}&#10;\\\\_nC_k= 2*3&#10;\\\\_nC_k=6<span>
</span>
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