Sali throws an ordinary fair 6 sided dice once.

a) what is the probability she gets a 6? /

Tim throws the same dice twice.

b) what is the probability he gets a 6 on both throws? /

Tony throws the dice once. He also throws a coin to get Heads or Tails. Two of the outcomes are 1H 2H

c) List all the other outcomes in the same format. 1H 2H?

a) 1/6

b) 1/36

c)

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

Step-by-step explanation:

a)

The probability of a certain event A to occur is given by

where

a is the number of successfull outcomes, in which event A occurs

n is the total number of possible outcomes

In this problem, the event is

"getting a 6 when throwing a dice once"

We know that the possible outcomes of a dice are six: 1, 2, 3, 4, 5, 6, so we he have

The successfull outcome in this case is only if we get a 6, so only 1 outcome, therefore

So, the probability of this event is

b)

In this case instead, we are throwing the dice twice.

The two throws of the dice are independent events (one does not depend on the other): the probability that two independent events A and B occur at the same time is given by the product of the individual probabilities,

where

is the probability that event A occurs

is the probability that event B occurs

Here we have:

- Event A is "getting a 6 in the first throw of the dice". We already calculated this probability in part a), and it is

- Event B is "getting a 6 in the second throw of the dice". Since the dice has not changed, the probability is still the same, so

Therefore, the probability of getting a 6 on both throws is:

c)

In this problem, we have:

- A dice that is thrown once

- A coin that is also thrown once

The dice has 6 possible outcomes, as we stated in part a):

1, 2, 3, 4, 5, 6

While the coin has two possible outcomes:

H = head

T = tail

So, in order to find all the outcomes of the two events combined, we have to combine all the outcomes of the dice with all the outcomes of the coin.

Doing so, using the following notation:

1H (getting 1 with the dice, and head with the coin)

The possible outcomes are:

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

So, we have a total of 12 possible outcomes.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

a. (14,375+570)(5.3) b. 158,417$

step-by-step explanation:

a. you are basically just setting up the equation. f= amount of debt hence 14,375. g= how much she is borrowing from her parents weekly hence 570 and t= the interest on her debt. creating the equation.

b.   14,375+570= 14,945

      14,945 x 5.3 = 79,208.5

      79,208.5 x 2(years) = 158,417(answer)