The first thing to do is to calculate how many ways you can choose 3 people from a set of eight. In order to do this, we need to use the attached formula.
(The letter 'n' stands for the entire set and 'r' stands for the number of objects we wish to choose.)
So we wish to choose 3 people ('r') form a set of 8 ('n')
combinations = n! / r! * (n - r)!
combinations = 8 ! / (3! * 5!)
combinations = 8 * 7 * 6 * 5! / (3!) * (5!)
combinations = 8 * 7 * 6 / 3 * 2
combinations = 56
Now of those 56 combinations, the 3 people can finish in 6 different ways.
For example, persons A, B and C could finish
ABC or ACB or BAC or BCA or CAB or CBA
So to get the TOTAL combinations we multiply 56 * 6 which equals
336 so the answer is (a)
Answer:
lies between 163.245 and 199.975
Step-by-step explanation:
Given
2 digit = 4#
Required
The range of
Let
--- the smallest possible value of #
So:
Let
--- the largest possible value of #
So:
<em>Hence, </em><em> lies between 163.245 and 199.975</em>
Answer:
1/2
Step-by-step explanation:
A regular die has 3 odd numbers (1, 3, and 5) and 3 even numbers (2, 4, and 6). The question asks for the probability that an odd number will not be rolled. This is the same as asking the probability of an even number being rolled. Since we have 6 numbers in our data set and 3 are even, we can deduce that 1/2 of them are even and the probability of an even number being rolled is 1/2.
HTH :)
The answer to the question is D
