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Kisachek [45]
3 years ago
6

Help ASAP Please I’ll give Brainly award

Mathematics
2 answers:
Sidana [21]3 years ago
4 0
The answer is 176 bro...good luck
Natali5045456 [20]3 years ago
3 0

Answer:

176

Step-by-step explanation:

trust

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What is the area of the figure below
Viefleur [7K]

Answer:

390cm

Step-by-step explanation:

1. find area of rectangle

 1. 15cm x 20cm= 300 cm

2. find area of triangle

  1. 1/2 (12 x 15)= 90cm

300+90

4 0
3 years ago
What is the simplified form of 30 times x to the sixth power over 14 times y to the fifth power times the fraction 7 times y-squ
Sonbull [250]
If we convert the given in its mathematical form, we have,

                      (30x⁶/14y⁵)(7y²/6x⁴)
It can be observed that the numerator of the first and the denominator of the second have a common factor of 6x⁴. Also, the denominator of the first and the numerator of the second expression have a common factor of 7y².

                      ((6x⁴)(5x²)/(7y²)(2y³))(7y²/6x⁴)

Cancellation of the common terms will give us an answer of,
  
                               <em>5x²/2y³

</em>
<em />Therefore, the simplified version of the involved operation is 5x²/2y³. <em>
</em>
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
“I think of a number and divide it by 5. The answer is 20.”
muminat

Answer: 100

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Free points, will give brainllest too
kow [346]

Answer: potato bread the new banana bread!

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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