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natali 33 [55]
3 years ago
10

PLEASE HELLP ME! I would appreciate if you can!!

Mathematics
1 answer:
S_A_V [24]3 years ago
3 0

Answer:

(2) A

Step-by-step explanation: easy Math

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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
Maura writes an expression equivalent to 8a - (-3 - 5a) - 2a. Her work is in the picture. In what step does Maura make a mistake
NISA [10]

Answer:

11a + 3

Step-by-step explanation:

Step 1 : 8a + 3 - 5a - 2a

error => - 5a

8a - (- 3 - 5a) - 2a =

= 8a + 3 + 5a - 2a

= (8a + 5a - 2a) + 3

= 11a + 3

7 0
3 years ago
Read 2 more answers
Mixed number as a decimal 7 9/18
Lesechka [4]

Answer:

3.5

Step-by-step explanation:

7 0
3 years ago
The weight of oranges growing in an orchard is normally distributed with a mean weight of 3.5 oz. and a standard deviation of 0.
Andreas93 [3]

Answer:

(3 oz ; 4oz)

Step-by-step explanation:

Given that :

Mean = 3.5 oz

Standard deviation = 0.5 oz

According to the empirical rule ; 68% is 1 standard deviation from the mean.

Hence, the interval that would represent the middle 68% will be:

(3.5 oz - 0.5 oz) ; (3.5 oz + 0.5oz)

(3 oz ; 4oz)

7 0
3 years ago
Please help me out for a brainlist :)
Nadya [2.5K]

Answer:

1,2,3,2,2

Step-by-step explanation:

because standard form is ax

7 0
3 years ago
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