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Vsevolod [243]
3 years ago
12

Please help!! Geometry Question!

Mathematics
2 answers:
romanna [79]3 years ago
4 0
6)
3x+15 = 75
3x = 60
x = 20
answer is B.
7)
x = 1/2(x - 9 + 2x - 3)
x = 1/2(3x -12)
x = 3/2x - 6
x = 1.5x - 6
-0.5x = -6
      x = 12
answer is D

8)
3x+5 = 4x - 40
4x - 3x = 5+40
x = 45
answer is A
EleoNora [17]3 years ago
4 0
6.) 20

7.) not sure, i think d

8.) a


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If Tina cuts a lawn by herself, she can finish in 5 hours. If bill cuts the same lawn by himself it takes him two hours longer t
shtirl [24]
Tina can finish in 5 hours
Bill takes two hours longer than Tina.
5+2=7
Bill takes 7 hours to cut the lawn.
7+5=12
It takes both of them working together 12 hours. Hope this helped!
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Express the product of (6-2i)(-4+3i) in a+bi form
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Multiply through the parentheses:
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18 + 26i

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2 years ago
PLEASE HELP ME I WILL MAKE YOU BRAINLYEST<br> solve <br><br> y^2-100=0
Tatiana [17]

Answer:

y=10

Step-by-step explanation:

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3 years ago
You pick a card at random. Without putting the first card back, you
Inga [223]

First, we need the probability of picking an odd number.

There are 5 cards in total, and 3 odd cards (3, 5, and 7).

That means that the probability that we'll draw an odd card would be \frac{3}{5}.

Then, we have 4 cards left, and 2 even cards (4 and 6), meaning that the probability that we draw an even card will be \frac{2}{4} or \frac{1}{2}.

To find the probability that these would happen in consecutive draws, we just multiply the probabilities together.

\frac{3}{5}*\frac{1}{2}=\frac{3}{10} or 0.3.

To convert this into a percentage, we multiply the decimal by 100.

0.3*100=30.

So the probability of picking an odd number and then picking an even number is 30%.

Hope this helps!

8 0
3 years ago
Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre
Whitepunk [10]

Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

Answer:

P(three quarters given two are quarters) = 1/7

PROBLEM 2:

Again, this is conditional probability. To help count the ways, let's instead count the ways to *not* draw any dimes. That means you have 2 choices for the first coin, 2 choices for the second coin and 2 choices for the third coin.

So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

3 0
3 years ago
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