Help ASAP!!! 1) A deck of cards has 6 yellow, 6 green, and 3 orange cards. You pick 3 cards from the deck. Each card is returned to the deck after each one is chosen. What is the probability that you choose 3 orange cards in a row?

2) Examine the contents of the marble bag. What is the probability of choosing a red marble, blue marble, and green marble in that order? Each pick is not replaced. THE FIRST IMAGE IS FOR THIS QUESTION.

3) Each of the spinners shown are spun once. What is the probability that the first spinner will land on either letter A or B and the second spinner will land on an even number? THE SECOND IMAGE IS FOR THIS QUESTION.

1) 3%

Step-by-step explanation:

1) 1/125

2) 1/15

3) 1/6

Step-by-step explanation:

Total Cards = 6 + 6 + 3 = 15

Orange cards: 3

1) Chance of first card orange:

3/15, there are 3 orange, 15 total

Chance of second card orange:

3/15, since it is returned, there is the same chance, and the third the same

3/15 * 3/15 * 3/15 = 1/5 * 1/5 * 1/5 = 1/125

2) There are 6 total, 2 green, 2 blue, and 2 red.

The chance of getting a red marble first is 2/6, or 1/3,  6 marbles, 2 red.

The chance of getting blue second is  2/5, since picks aren't replaced.

Chance of getting green third is 2/4, or 1/2, since picks aren't replaced

1/3 * 2/5 * 1/2 = 2/30, or 1/15

3) First spinner has 6 total, 2 are A or B, so 2/6 or 1/3.

Second spinner has 4 total, 2 evens, so 2/4 or 1/2

Now multiply 1/3 * 1/2 = 1/6

step-by-step explanation:

for the free

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.