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alexdok [17]
2 years ago
5

Please show me how you got the answer. Will give brainliest.

Mathematics
1 answer:
umka21 [38]2 years ago
4 0

Answer:          410 possible points

To solve for the possible points in the course, use a proportion and cross multiply.

<h2>What is a proportion?</h2>

A proportion is two fractions that are equal to each other. For the fractions to be equal, they are proportionate. In our math problem, percentage is proportionate the the number of points.

The fractions represent information you have, and one variable will represent what you are solving for.

<h2 /><h3>Given</h3>

Your points = 361

Points to get an 'A' = your points + 8

'A' = 90% of possible points

<h3>Find the points needed for an 'A' at 90%</h3>

Points to get an 'A' = your points + 8

= 361 + 8

= 369

<h3>State variables</h3>

let <em>x</em> be the amount of possible points

<h3 /><h3>Write a proportion</h3>

We know the number of points for 90% is 369 points. Remember that 90% is the same as 90/100.

Keep the information with the same units inside the same fraction. Here, percentage is represented by the first fraction. Points is represented by the second fraction.

\displaystyle{\frac{90}{100}=\frac{369}{x}}

<h3>Solve</h3>

We can solve using cross multiplication. This is when we multiply each numerator by the denominator from the other fraction.

90x = 369*100        Simplify by multiplying.

90x = 36,900            Isolate the variable <em>x</em>.

\displaystyle{\frac{90x}{90} = \frac{36,900}{90}}           Divide both sides by 90 to cancel out 90 on the left.

\displaystyle{x = \frac{36,900}{90}}              Divide.

x = 410                    Solved for <em>x</em>, representing the possible points.

∴ The possible points in the art course is 410.

Learn more about finding a missing number given percentage here: brainly.com/question/26535441

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Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

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Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

                                                                   =  \frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}

                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

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Step 1 : Simplify both sides
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Hey there!
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Let's try the third one:

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