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:
$(8967/n + 8.4)
Step-by-step explanation:
0.4nx + 0.6n(x+21) = 8967
nx + 12.6n = 8967
x = 8967/n - 12.6
x+21 = 8967/n + 8.4
Where n is the no. of balls
Example: if total balls were 300
n = 300
More expensive one would cost:
8967/300 + 8.4 = $38.29
Answer:
At price 3 and 11, the profit will be $0
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one.
<em>
A certain companies main source of income is a mobile app. The companies annual profit (in millions of dollars) as a function of the app’s price (in dollars) is modeled by P(x)=-2(x-3)(x-11) which app prices will result in $0 annual profit?</em>
My answer:
Given:
- x is the app price
- P(x) is the profit earned
If we want to find out the app price that will result in $0 annual profit? It means we need to set the function:
P(x)=-2(x-3)(x-11) = 0
<=> (x-3)(x-11)= 0
<=> x - 3 = 0 or x - 11=0
<=> x = 3 or x = 11
So at price 3 and 11, the profit will be $0
Hope it will find you well.
<span> Honor roll Not on honor roll Total
Received math class requested 315 64 379
Did not get math class requested 41 80 121
Total 356 144 500
Honor roll: request granted: 315/356 = 0.88 x 100% = 88%
Not Honor roll request granted: 64/144 = 0.44 x 100% = 44%
Honor roll students were given preference in granting request than those not in the honor roll.</span>
Answer:
Step-by-step explanation:
Given that in the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College.
Table is prepared as follows
Bus coll. Lib Arts coll Educ. coll Total
Observed 90 120 90 300
Expected p.c. 35 35 30 100
Expected number 105 105 90 300
Percent*300/100
Hence expected frequency for business college = 105
