By finding the probability, we can expect that 2 out of the 30 students will win the prize.
<h3>
How many of the 30 students will win the prize?</h3>
First, we should get the probability of winning this game.
Here the students select two numbers out of 6, just to make the calculations let's say that these numbers are 1 and 2.
Now, we need to get these two numbers in two balls. The probability of getting the ball with the number 1 out of the 6 balls is:
p = 1/6
The probability of getting the ball with the number 2 out of the remaining 5 balls (because we already got one) is:
q = 1/5
The joint probability is then:
P = 2*(1/6)*(1/5) = 1/15.
Where the factor 2 comes to take in account the permutations, for the case where we first draw the number 2 and then the number 1.
Then the expected number of students that will win is equal to the probability times the total number of students:
(1/15)*30 = 2
So out of the 30 students, we can expect that 2 will win.
If you want to learn more about probability:
brainly.com/question/25870256
#SPJ1
76, 84, 93, 67, 82, 87, & 76
The mean is just the middle number of the data set,
Start by rewriting the set in order from smallest to biggest:
67, 76, 76, 82, 84, 87, 93
Determine which is the number in the middle,
67, 76, 76, *82*, 84, 87, 93
So, in this case, the median is 82.
I would pick C sorry if it’s wrong
Answer:
Let the Dulcina's collection be 'x'
Let the Tremaine collection be 'x-39'
x + x - 39 =129
2x = 129 +39
2x = 168
x = 168/2
x = 84
Dulcina's collection = x = 84
Tremaine's collection = x - 39 = 84 - 39 = 45
37 is 26%(rounded up to the nearest percent) of 144
