Answer:
a. Yes(n=500>=5, n(1-p)=25>=5)
b. 0.15241
Step-by-step explanation:
a. A normal approximation to the binomial can be used
5 and n(1-p)>=5:
#We calculate our p as follows:
=x/n=470/500=0.94
n=500
n(1-p)=500(1-0.95)=25
Hence, we can use the normal approximation.
b. This is a normal approximation.
-Given that p=0.95(95%)
-We verify if our distribution can be approximated to a normal:

Hence, we can use the normal approximation of the form:

Hence, the probability of the sample proportion is the same as the proportion of the sample found is 0.15241
C=Pi X D
Diameter is twice the radius so that would be 24 feet
C = Pi X 24
C = (22/7) X 24
C = 75.428 Feet.
Given:

To find:
The highest and lowest scores Sam could have made in the tournament.
Solution:
We have,


It can be written as

Add 288 on both sides.

and 
and 
Therefore, the highest and lowest scores Sam could have made in the tournament are 290 and 286 respectively.
Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is
As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is
. Thus, there are only 35 ways to distribute the blackboards in this case.
Answer:
x=9
Step-by-step explanation:
Let total bundle=x
Morning edition of the daily sun=1/3x
=x/3
Afternoon edition of the daily sun=2
That leaves 2x/3 - 2, of which
x/3 - 1 are regional
1 is local
1 from another state
Sum everything
x = x/3 + 2 + x/3 - 1 + 1 + 1
x=2x/3 +3
x-2x/3=3
3x-2x/3=3
x/3=3
Cross product
x=3×3
x = 9