Answer:
608
Step-by-step explanation:
There are a total of 1,824 drinks since there were 76 cases at 24 drinks a piece. This creates a total number of 24*76 = 1824.
If each student receives 3 then 1824/3 = 608. There were 608 students.
Answer: A) 20.9 ; B) 34years
Step-by-step explanation:
Given the following :
AGE (X) - - - - - - - 19 - -20 - - - 21 - - - 22 - - - 23
FREQUENCY (F) - 2 - - 3 - - - - 1 - - - - 4 - - - - 1
A)
MEAN(X) = [AGE(X) × FREQUENCY (F)] ÷ SUM OF FREQUENCY
F*X = [(19 * 2) + (20 * 3) + ( 21 * 1)+(22 * 4)+(23 * 1)]
= 38 + 60 +21 + 88 + 23 = 230
SUM OF FREQUENCY = 2 + 3 + 1 + 4 + 1= 11
MEAN(X) = 230 / 11
X = 20.9
B)
WHEN A NEW PLAYER WAS ADDED :
MEAN (X) = 22
Let age of new player = y
Sum of Ages = 19 + 19 +20 + 20 + 20 + 21 + 22 + 22 + 22 + 22 + 23 + y
Number of players = 11 + 1 = 12
Mean(x) = sum of ages / number of players
New mean (x) = 22
x = (230 + y) / 12
22 = (230 + y) / 12
Cross multiply
264 = 230 + y
y = 264 - 230
y = 34 years
Answer:
12.25
Explanation
49/20 <em><u>(which is the same as (</u></em><em><u>4</u></em><em><u>9</u></em><em><u>÷</u></em><em><u>2</u></em><em><u>0</u></em><em><u>))</u></em><em><u> </u></em>=2.45
<em><u>Multiply</u></em><em><u> </u></em><em><u>2</u></em><em><u>.</u></em><em><u>4</u></em><em><u>5</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>amount</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>hours</u></em><em><u> </u></em><em><u>(</u></em><em><u>5</u></em><em><u>)</u></em><em><u> </u></em>

=12.25
Answer:
See the proof below.
Step-by-step explanation:
For this case we just need to apply properties of expected value. We know that the estimator is given by:

And we want to proof that 
So we can begin with this:

And we can distribute the expected value into the temrs like this:

And we know that the expected value for the estimator of the variance s is
, or in other way
so if we apply this property here we have:

And we know that
so using this we can take common factor like this:

And then we see that the pooled variance is an unbiased estimator for the population variance when we have two population with the same variance.
A) -2.8 > -3.5 is true. There are infinite solutions.
B) -7.75 < -8.45 is false. There is no solution.
C) 6.52 < 6.5 is is false. There is no solution.
D) -11.1 > 10.7 is false. There is no solution.