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3 years ago

In a small private school three students are randomly selected from 18 available students what is the probability that

Mathematics

1 answer:

Komok [63]3 years ago

3 0

1/ 816

Step-by-step explanation:

Step 1 :

Given

Total number of students in the school = 18

Number of students randomly selected = 3

Step 2 :

Number of ways in which 3 students can be selected out of 18 students is

C(18,3) = 18! /(3!(*18-3)!) = 18*17*16/3*2*1 = 816

Number of ways in which 3 students are selected and they are the youngest = 1

Hence the probability of selecting 3 youngest students from the available students is 1/ C(18,3) = 1/ 816

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Roy Gross is considering an investment that pays 7.6 percent, compounded annually. How much will he have to invest today so that

Debora [2.8K]

Answer: he should invest $16129 today.

Step-by-step explanation:

Let $P represent the initial amount that should be invested today. It means that principal,

P = $P

It would be compounded annually. This means that it would be compounded once in a year. So

n = 1

The rate at which the principal would be compounded is 7.6%. So

r = 7.6/100 = 0.076

The duration of the investment would be 6 years. So

t = 6

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years.

A = 25000

Therefore

25000 = P(1+0.076/1)^1×6

25000 = P(1.076)^6

25000 = 1.55P

P = 25000/1.55

P = $16129

4 0

3 years ago

If the numbers 4, 5 and 6 are each used exactly once to replace the letters in the expression A ( B − C ), what is the least pos

guapka [62]

Answer:

The least possible result is -10.

Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

First of all,

Let A = 4, B = 5 and C = 6

$A(B-C) = 4 \times (5-6) = 4 \times -1 = -4$

Let A = 4, B = 6 and C = 5

$A(B-C) = 4 \times (6-5) = 4 \times 1 = 4$

Let A = 5, B =4 and C = 6

$A(B-C) = 5 \times (4-6) = 5 \times -2 = -10$

Let A = 5, B = 6 and C = 4

$A(B-C) = 4 \times (6-4) = 4 \times 2 = 8$

Let A = 6, B = 4 and C = 5

$A(B-C) = 6 \times (4-5) = 6 \times -1 = -6$

Let A = 6, B = 5 and C = 4

$A(B-C) = 6 \times (6-5) = 6 \times 1 = 6$

Summarizing the above values in the form of a table:

$\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\ 4 & 6 & 5 & 4\\ 5 & 4 & 6 & -10\\ 5 & 6 & 4 & 10\\ 6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}$

So, the least possible result is -10.

3 0

3 years ago

What is 6C3? As in, how many ways can you choose 3 objects out of 6 objects, where order DOES NOT matter?

marysya [2.9K]

You could find this by doing 6 * 5 * 4, or 6 choices * 5 choices * 4 choices = total choice combinations.
6*5*4 = 120

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3 years ago

What do they call the international, hula hoops championship? Its for a math worksheet

lapo4ka [179]

They call it the Whirl Series ^.^

7 0

3 years ago

I. EED HELP FAST, IS FOR A TEST!!!!

elixir [45]

Answer:

29%

Step-by-step explanation:

I guarantee this is the right answer.

4 0

2 years ago