Start by letting x represent the number of Sam's pencils. Then Sari has 3x (since she has three times as many)
Together they total 28 pencils:
x + 3x = 28
4x = 28 /:4 (divide both sides by 4)
x = 7
So Sam has 7 pencils.
Sari, having three times as many, has 7 * 3 = 21.[[TAG_8]]
Answer:
a) Ann has a 1/3 chance of winning in the first round
b) The chance of Ann winning for the first time in the fourth round is 8/81
c) The probability that Ann's first win occurs after the fourth round is 16/81
Step-by-step explanation:
a) Each strategy is played with a probability of 1/3. Given any strategy, there’s a 1/3 chance that Bill will choose the strategy that allows Ann to win. Consequently, the probability of Ann securing a victory in the first round (or any round) is
1/3 * 1/3 + 1/3 * 1/3 + 1/3 * 1/3 = 1/9 + 1/9 + 1/9 = 1/3.
Thus, the likelihood of Ann winning the initial round is 1/3.
b) The chances of Ann winning a round stand at 1/3; therefore, her chances of not winning are 2/3. This must happen three times before her first victory. Thus, the probability that Ann's first win occurs in the fourth round is
(2/3)³ * 1/3 = 8/81.
c) The first victory happens after the fourth round if she remains unsuccessful in the first four rounds, translating to a possibility of (2/3)⁴ = 16/81.
Y = 3bx - 7x
y = x(3b - 7)
Assuming 3b - 7 ≠ 0, divide both sides by 3b - 7.

Solution:
Answer: Indeed, because the input value ranges from 0 to 12.
Step-by-step explanation:
To address this query, we first need to formulate a function that reflects the situation.
Every student engaging in community service earns the club $5
E(s) = 5 s
Where:
- E= total earnings
- n = number of students participating in community services (input value)
Indeed, since the input value n indicates how many students from the club engage in community service, and with the total number of students being 12, the input can indeed vary between 0 and 12.