Question

can someone check if I did the ones I did right and help me on the ones I didn't do?

Answer 1

31. and 35. are correct.

Answer 2

okay thx can u help me on the ones i didnt do?

Answer 3

For #32 try to distribute then simplify.

Answer 4

Yes, for 32. use the distributive property twice. Then combine like terms.

Answer 5

Do you need help with that?

Answer 6

yes

Answer 7

Here is problem 32.

\(\large 2xy^2(3x^2 - 2y) - x^2y(2x - 3xy) \)

Answer 8

We need to distribute 2xy^2 through the first set of parentheses, and x^2y through the second set of parentheses.

Answer 9

oh is it c?

Answer 10

\(\large =2xy^2(3x^2) - 2xy^2(2y) - x^2y(2x) + x^2y(3xy)\)

\(\large =6x^3y^2 - 4xy^3 - 2x^3y + 3x^3 y^2\)

\(\large = 9x^3 y^2 - 4xy^3 - 2x^3 y \)

Answer 11

Now it's completely done.
Compare the answer with the choices.

Answer 12

i oh what i did wrong i subtracted

Answer 13

Ready for 33.?

Answer 14

yes

Answer 15

You need to find a probability of one event following another event.
You need to find each individual probability and then multiply them together.

Answer 16

\(probability ~of ~event = \dfrac{number ~of~desired~outcomes}{total~number~of ~outcomes} \)

Answer 17

The first event is the first drawing.
You have a bag with marbles, and you want to draw a red marble.
The desired outcome is a red marble.
How many red marbles are there?

Answer 18

3/16

Answer 19

Right. For the first drawing you have 3 marbles out of 16, so the the probability of picking a red marble the first time is 3/16.

Answer 20

Once you picked the first marble, and it was red, you do not replace it.
Now you have the second event.
Again, you want to pick a red marble out all the marbles left.
For the second drawing, how many red marbles are there, and how many total marbles are there?

Answer 21

that's the part i am a bit confused

Answer 22

Ok.
Remember the problem tells you there is no replacement.
This is a very important piece of information.
That means that once you draw the first marble, and it is red, you now have one less marble left.
That means for the second drawing, you are now interested in 2 red marbles out of 15.
What is the probability of drawing a red marble out of a bag that has 15 marbles, and only 2 are red?

Answer 23

2/16

Answer 24

Not 16 anymore.
Remember that once you drew the first red marble, there is one less red marble, so only 2 red marbles, and also one less marble in total, so only 15.

Answer 25

oh i see what u mean by

Answer 26

so 2/15

Answer 27

Correct.
Now you have the probabilities of the two separate events.
3/16 and 2/15.
Two find the probability of both events happening, multiply the two individual probabilities.

Answer 28

so u want me to multiply them together now?

Answer 29

yes

Answer 30

\(p(red~marble~followed~by~red~marble) = \dfrac{3}{16} \times \dfrac{2}{15} \)

Answer 31

6/240=2/80=1/40

Answer 32

correct

Answer 33

Ready for 34.?

Answer 34

yes

Answer 35

A negative exponent in the numerator is a positive exponent in the denominator.