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Vesnalui [34]
2 years ago
7

20 points please help me i need this answer

Mathematics
1 answer:
lozanna [386]2 years ago
4 0

The statements that are TRUE regarding the box plots are:

A, D, and E.

The box plots are given in the image attached below.

<h3>The Five-number Summary Displayed by a Box Plot</h3>

The values displayed by a box plot for a given data are:

  • Median
  • Minimum
  • Maximum
  • First Quartile
  • Third Quartile

Interquartile range = Third Quartile - First Quartile.

Thus, shelter's A median is 21, shelter B's 18.

Interquartile range for shelter A = 28 - 8 = 20

Interquartile range for shelter B = 28 - 10 = 18

<h3>What is a Symmetric Data Set?</h3>
  • If a data set is symmetric, a box plot that represents the data set will have the median in the middle of the box, while both whiskers will be about the same on both sides of the box.
  • For example, the data set for shelter B is symmetric.

Therefore, the statements that are TRUE regarding the box plots are:

A, D, and E.

Learn more about box plots on:

brainly.com/question/14277132

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Step-by-step explanation:

Probability that Dan chooses a Green card = 0.2

Probability that Esther chooses a Green card = 0.2 (same set of cards = same probability)

Since they are independent events,

total probability = 0.2 * 0.2 = 0.04.

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7. What is the slope of the line that passes through (12, 10) and (11, 15)?
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Answer:

-5

Step-by-step explanation:

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What is the solution to the inequality below 8 m&gt; 40
nalin [4]

Answer:

M>5

Step-by-step explanation:

Lets just rewrite the equation

8m>40 is the same as m(8)>40

lets divide

8m / 8>40/8

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What is the domain of the function f(x)=√6−2x?
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4 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

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3 years ago
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