Please help? ———-
1. A coin is weighted so that it lands on “heads” 40% of the time. Find the probabilities of getting 0, 1, 2, 3, and 4 heads when you flip the coin 4 times.

2.Using the weighted coin above, you play a game. You pay $2 to play. If you get 0, 1, or 2 heads, you lose. If you get 3 heads, get your money back. If you get 4 heads, you win $10. What are your expected winnings?

question:

could we have options?

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation:

an = 22 + (n-1) *6

110 seats

step-by-step explanation:

an=a1+(n-1)(d)

a1 = 22   this is the first item in the sequence

d =6   this is how the sequence goes up

a)   an = 22 + (n-1) *6

b) a12 = 22 + (12-1) *6

    a12 = 22+11*6

  a12 = 22 +88

  a12 = 110

answer: c

substitute in the values.

for a:

f(0) = 0 + 9

f(0) = 9

a is a solution.

for b:

f(7) = 14 + 9

f(7) = 23

b is a solution.

for c:

f (3) = 6 + 9

f(3) = 15

c is not a solution. it's supposed to be 14.

for d:

f(2) = 4 + 9

f(2)= 13

d is a solution.