2) Suppose 120 people are surveyed and asked about their eating habits. 50 people stated that they like Raising Cane's chicken fingers, 80 people said that they like

Layne's chicken fingers. If 10 people don't like either, then...

a) What is the probability that someone surveyed likes both Raising Cane's and

Layne's chicken fingers?

1/6

Step-by-step explanation:

Given that :

Total surveyed (u) = 120

n(Cane's chicken fingers) = 50

n(Layne's chicken fingers) = 80

n(do not like either) = 10

a)

What is the probability that someone surveyed likes both Raising Cane's and Layne's chicken fingers?

n(Cane's n Layne's) = 20

u = 120

P = required outcome / total possible outcomes

P(Layne's and Cane's chicken) = [n(Cane's n Layne's)/ u]

= 20 / 120

= 1/6

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6