We have been given in a cohort of 35 graduating students, there are three different prizes to be awarded. We are asked that in how many different ways could the prizes be awarded, if no student can receive more than one prize.
To solve this problem we will use permutations.
We know that formula for permutations is given as
On substituting the given values in the formula we get,
Therefore, there are 39270 ways in which prizes can be awarded.
Answer:
20,800
Step-by-step explanation:
Answer:
just breather just breathe
Step-by-step explanation:
Answer:
0.98046
Step-by-step explanation:
Given:
Here we are required to find
P(5.48 <X<5.82) and between 5.48 and 5.82 we each z-score given by
<em>z = (x - μ) / σ</em>
So 5.48 we have z = -2.714 and 5.82 we have z = 2.143
Therefore we have the area of interest on the normal distribution chart given by
:
P( -2,714 < z < 2.143)
= 1 - P(Z<-2.714) + P(Z > 2.143)
= 1 - P(Z<-2.714) + (1 - P(Z < 2.143))
= 1 - 0.00336 + 1 - 0.98382
= 1 - 0.01954 = 0.98046
