Answer:
<h2>a) 1.308*10¹² ways</h2><h2>b)
455 way</h2>
Step-by-step explanation:
If there are 15 balls labeled 1 through 15 in a standard football game, the order of arrangement of the 15 balls can be done in 15! ways.
15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2
15! = 1.308*10¹² ways
b) If 3 of the 15 balls are to be chosen if order does not matter, this can be done in 15C3 number of ways. Since we are selecting some balls out of the total number of balls, we will use the concept of combination.
Using the combination formula nCr = n!/(n-r)!r!
15C3 = 15!/(15-3)!3!
15C3 = 15!/12!3!
15C3 = 15*14*13*12!/12!*6
15C3 = 15*14*13/6
15C3 = 455 ways
if the number of times spun is not a multiple of 8, it is impossible to land on an exactly even distribution for this spinner. But since the probabilities of landing on an exactly even distribution are so small, that detail doesn't perturb the overall trend, except perhaps down at n=8 spins So group 3
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.
