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malfutka [58]
2 years ago
12

I need helpppppppppppp

Mathematics
2 answers:
Elodia [21]2 years ago
6 0

Answer:1/4 I think

Step-by-step explanation:3/4-1/2=1/4

kakasveta [241]2 years ago
4 0

Answer:

1/4 I'm pretty sure

Step-by-step explanation:

because half of the class already went to the afternoon session and 3/4 went to see the movie

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The number of people attending the first basketball game of the season was 840. The number of people attending the last game of
melamori03 [73]

Answer:

Step-by-step explanation:

percent increase = (new number - original number) / (original number) * 100

                             = (1200 - 840) / (840) * 100

                             = (360) / 840 * 100

                              = 0.4285 * 100

                              = 42.85% rounded to nearest percent = 43% <==

5 0
3 years ago
The sum of one and the product of a number and two is thirteen
Zigmanuir [339]

Answer:

2x + 1=13

Step-by-step explanation:

the product of a number and 2 (2x)

sum of 1 (+1)

is 13 (=13)

6 0
3 years ago
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
Write the decimal equivalent to 7/9
diamong [38]
Just do 7 divided by 9

3 0
3 years ago
Read 2 more answers
Select the multiplication equation that could represent the following question: how many 3/8s are in 5/4
muminat

Answer:

The answer is: 3 1/3.

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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