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Len [333]
3 years ago
9

Rude took a test and earned a 92%. Camilo earned an 88% on the same test. If Rudy answered 46 questions correctly, how many more

questions did he answer correctly than Camilo?
Mathematics
1 answer:
g100num [7]3 years ago
7 0
Rudy answered 2 more questions correctly
...........
46x2=92
44x2=88
...........
92-88=4
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An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0
3 years ago
This figure is 14 of a circle.
Wittaler [7]

Answer:

The perimeter of a quarter of a circle with radius 2 inches is 7.1 inches.

Step-by-step explanation:

Given : A quarter of a circle with radius labeled 2 in.

Quarter of circle means fourth part of a circle  (as shown in image by shaded part).

We have to find the perimeter of this shaded quarter of circle.

Perimeter is the sum of the dimension of the given figure.

Perimeter of circle is 2\pi r

Since, we are given Quarter circle so its perimeter \frac{2}{4} \pi r+r+r=\frac{1}{2} \pi r+r+r

Perimeter of given figure = \frac{1}{2} \pi r+r+r

Perimeter of given figure = \frac{1}{2}\cdot 3.14 \cdot 2+2+2=3.14+4=7.14

Thus, the perimeter of a quarter of a circle with radius 2 inches is 7.1 inches.

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Anny goes bowling. He has $25.00 to spend. He spends $4.25 to rent shoes. He spends $2.50 for each game he bowls. Which inequali
spayn [35]

Answer:

D. 25 +4.25x < 25

Step-by-step explanation:

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Did you hear about the two punsters who told a lot of jokes about cats
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Cool. and. gooooooooood.
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3 years ago
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What percentage is 15 out of 17?
Ket [755]
88.23%

Simply plug 15/17 into you calculator 

and it will give you a decimal, in this case it should be .88235294...

simply move the decimal back to placements and you should have

88.235294... thats still a lot of numbers so just reduce it to 2 places after the decimal and add a percent sign

88.23%

and your done :) this works for any similar problems

I hope this helps

8 0
3 years ago
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