No there is not. 50 divided by 1 2/3 gives you 30 which means it can’t create 35 servings.
The expected value should be a 1/3 change because you have three coins each , then 1/6 altogether change or randomization the most realistic highest expected value is you paying them 2$
Answer:
(-5/9, 16/9)
Step-by-step explanation:
2y = -x + 3
-y = 5x + 1
To find the intersection, you need to substitute the y-value from the second equation into the first equation. Rearrange the second equation so that it is equal to y.
-y = 5x + 1
-1(-y) = -1(5x + 1)
y = -5x - 1
Substitute this equation into the y-value of the first equation.
2y = -x + 3
2(-5x - 1) = -x + 3
-10x - 2 = -x + 3
(-10x - 2) + 2 = (-x + 3) + 2
-10x = -x + 5
(-10x) + x = (-x + 5) + x
-9x = 5
(-9x)/(-9) = (5)/(-9)
x = -5/9
Plug this x value into one of the equations and solve for y.
2y = -x + 3
2y = -(-5/9) + 3
2y = 5/9 + 3
2y = 32/9
(2y)/2 = (32/9)/2
y = 32/18 = 16/9
The ordered pair is (-5/9, 16/9).
Answer:
The absolute maximum on the interval 0 < x < 3 is at x = 2 and f(2) = 0. Since x = 2 can only give an absolute maximum, so there is no absolute minimum.
Step-by-step explanation:
f(x) = (-x + 2)⁴
to find the absolute maximum and minimum values, we differentiate f(x) with respect to x.
So df(x)/dx = f'(x) = 4(-x + 2)³
The maximum and minimum values are obtained when f'(x) = 0
So, 4(-x + 2)³ = 0
⇒ (-x + 2)³ = 0
⇒ -x + 2 = 0
-x = -2
x = 2
Now, f(2) = (-2 + 2)⁴ = 0⁴ = 0
So, the absolute maximum on the interval 0 < x < 3 is at x = 2 and f(2) = 0. Since x = 2 can only give an absolute maximum, so there is no absolute minimum.
So to make an inequality, and we let c=number of coins then:
c≥6