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laiz [17]
3 years ago
13

Will give brainliest

Mathematics
2 answers:
sweet [91]3 years ago
6 0

Answer:

B

Step-by-step explanation:

I did this test and got this one correct.

nalin [4]3 years ago
5 0

Answer:

give me brainliest first

Step-by-step explanation:

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TRUE

Step-by-step explanation:

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2 years ago
Simplify.<br> 7x² - x + 9<br> + 6x² + 5x - 10<br> ___________<br><br> Will give brainliest.
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Answer:

13²+4−1

Step-by-step explanation:

<em><u>hope it helps</u></em>

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2 years ago
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Students on a field trip at an amusement park were asked their grade in school. The table shows the results of the survey.
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Can you graph a line if its slope is undefined? Explain.
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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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