A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Answer: The mean and variance of Y are $0.25 and $6.19 respectively.

Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)=

You win $6 if both faces are heads, so P(Y=6)=P(HH)=

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)=

To find the expected value to win: E(Y)=

Thus, the mean of Y: E(Y)= $0.25

Variance =

Therefore, variance of Y = $ 6.19

Answer:

Approximately 59 stacked cups.

Step-by-step explanation:

We have the following measurements:

Height of one cup = 12.5 cm,

Height of two cups stacked = 14 cm,

Height of three cups stacked = 15.5 cm,

...and so on.

This situation can be described by an arithmetic sequence,

12.5, 14, 15.5,....

The first term is defined as a = 12.5,

with a common difference of d = 1.5 cm.

Thus, the height of x stacked cups is given by

As per the problem,

h(x) = 200

⇒ 1.5x + 11 = 200

⇒ 1.5x = 189

⇒ x = 59.3333333333 ≈ 59.

Therefore, you will need approximately 59 stacked cups.

Result:

0.884

Detailed steps:

Initially, calculate the exponents, perform the foil, combine like terms, and the resulting figure will be 0.884.