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mr_godi [17]
3 years ago
15

What is the probability that a point chosen at random in the given figure will be inside the smaller square?

Mathematics
2 answers:
UNO [17]3 years ago
3 0
Your anwser should be
9/25
Aloiza [94]3 years ago
3 0
They are right! The answer is 9/15! :3

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If one number is the multiple of another number, then the L.C.M. will be the smaller number (the number whose multiple the other number is).

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You flip a coin eight times. find the number of possible outcomes in the sample space.
nika2105 [10]
2( coin faces) 8( number of times)
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3 3/8 divided by 9 =
wlad13 [49]

Answer:

0.375

Hope this helps....

Have a nice day!!!!

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Emily went to the movie theatre for her birthday. A mix of adults and children attended, making a total of 19 people. Each adult
konstantin123 [22]

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13 Adults and 6 Kids

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AURORKA [14]

Answer:

1)

Minimum is a 4th Degree

2)

Positive; Even

3)

x=-4, -1\text{ Odd Multiplicity}\\x=3\text{ Even Multiplicity}

Step-by-step explanation:

Part 1)

The minimum degree of our function will be 4.

Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.

So, it can be anything 1, 3, 5, 7, etc.

However, we will choose the lowest one, 1.

Next, we know that the graph <em>bounces off</em> at 3.

So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.

We choose the lowest one, 2.

Therefore, the minimum degree of our function will be 1+1+2 or 4.

Part 2)

The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.

Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.

This applies to all polynomials with even degrees.

Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.

Part 3)

This is similar to Part 1.

We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is: x=-4, -1, 3

We know that it<em> passes through</em> x=-4 \text{ and } x=-1 . So, these two factors must have an odd multiplicity.

However, since the graph <em>bounces off</em> x=3, this factor must have an even multiplicity.

And we're done!

7 0
2 years ago
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