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3 years ago

Question 101 points)

Mathematics

1 answer:

maw [93]3 years ago

4 0

Answer:

$\large\boxed{y=-\dfrac{2}{3}x+\dfrac{13}{3}}$

Step-by-step explanation:

$\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\===============================$

$\text{We have the point:}\\\\(5,\ 1)\ \text{and}\ (-4,\ 7).\ \text{Substitute:}\\\\m=\dfrac{7-1}{-4-5}=\dfrac{6}{-9}=-\dfrac{6:3}{9:3}=-\dfrac{2}{3}\\\\\text{We have the equation in form:}\\\\y=-\dfrac{2}{3}x+b\\\\\text{Put the coordinates of the point (5, 1) to the equation:}\\\\1=-\dfrac{2}{3}(5)+b\\\\1=-\dfrac{10}{3}+b\qquad\text{add}\ \dfrac{10}{3}\ \text{to the both sides}\\\\\dfrac{3}{3}+\dfrac{10}{3}=b\to b=\dfrac{13}{3}\\\\\text{Finally:}\\\\y=-\dfrac{2}{3}x+\dfrac{13}{3}$

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If 6!+7!+8!=(n)(6!), then the value of n is<br> A) 15<br> B)16<br> C)63<br> D)64<br> E)15!

gladu [14]

6!+7!+8!=(n)(6!)
calculate the individual values first:
6!=720
7!=5040
8!=40320

plug them into the equation:
720+5040+40320=n720
solve for n
5760+40320=n720
46080=n720
divide both sides by 720 to isolate n
46080/720=n720/720
64=n

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3 years ago

A triangle has sides with lengths of 18 centimeters, 68 centimeters and 74 centimeters. Is it a right triangle?

Hoochie [10]

18^2 + 68^2 =

= 324 + 4,624

= 4,948

74^2 = 5,476

4,848 is not equal to 5,476.

Since the sum of the squares of the lengths of the two shorter sides is not equal to the square of the length of the longest side, the triangle is not a right triangle.

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3 years ago

List all the ways to select two members from S with repetition. The order in which the members are selected is not important. Fo

Taya2010 [7]

9514 1404 393

Answer:

KK KL KM KN KP LL LM LN LP MM MN MP NN NP PP

Step-by-step explanation:

Instead of starting with adjacent letters in the list, we start with the same letter. There are 15 possible pairs.

KK KL KM KN KP LL LM LN LP MM MN MP NN NP PP

<em>Additional comment</em>

We leave out pairs with letters previous, because you've said KL is the same as LK, and KL has already been listed. The problem with this list is that it misstates the probabilities of randomly obtaining a given pair. It is not 1/15, but is 2/25 for all but letter duplicates. For duplicates (KK), the probability is 1/25, not 1/15.

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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!

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

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

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.

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

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ 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