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Viktor [21]
3 years ago
12

Introduction to Functions quiz answers?

Mathematics
1 answer:
larisa86 [58]3 years ago
6 1

Answer:

hgdgcvwjhmcvtuqyjbwq jxghqnwukdhjqwnoukjxqexqe

Step-by-step explanation:

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Someone help 9th grade math i will give brainliest
DENIUS [597]

Answer:

I think it may be H. but let me know if i am wrong. if i am right plz mark brainliest

Step-by-step explanation:

27 divided by 3 is 9, 27/3=9 because 27/3 is the same exact thing as 27 divided by 3. So the answer has to be 9

6 0
3 years ago
What is the simplified expression for 6 (2 (y x))? 6 y 12 x 12 y 12 x 12 y 8 x 8 y 8 x
stellarik [79]

The simplified expression for 6(2( y + x)) is 12y + 12x.

Accordin to the given question.

We have an expression

6(2( y + x))

Therefore,

The simplified expression for 6(2( y + x)) is given by

6(2(y + x))

= 6(2y + 2x)          (by distributive law)

= 12y + 12x              (by distributive law)

Hence, the simplified expression for 6(2( y + x)) is 12y + 12x.

Find out more information about simplified expression here:

brainly.com/question/18077352

#SPJ4

5 0
1 year ago
What is 45% of 649926391​
MArishka [77]

Answer:

292,466,875.95

Step-by-step explanation:

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
How many lb of brand X sugar which
Dafna1 [17]

Answer:  <u>4 pounds</u> of brand X sugar

====================================================

Reason:

n = number of pounds of brand X sugar

5n = cost of buying those n pounds, at $5 per pound

Brand Y costs $2 per pound, and you buy 8 lbs of it, so that's another 2*8 = 16 dollars.

5n+16 = total cost of brand X and brand Y combined

n+8 = total amount of sugar bought, in pounds

3(n+8) = total cost because we buy n+8 pounds at $3 per pound

The 5n+16 and 3(n+8) represent the same total cost.

Set them equal to each other. Solve for n.

5n+16 = 3(n+8)

5n+16 = 3n+24

5n-3n = 24-16

2n = 8

n = 8/2

n = 4 pounds of brand X sugar are needed

-------------

Check:

n = 4

5n = 5*4 = 20 dollars spent on brand X alone

16 dollars spent on brand Y mentioned earlier

20+16 = 36 dollars spent total

n+8 = 4+8 = 12 pounds of both types of sugar brands combined

3*12 = 36 dollars spent on both types of sugar brands

The answer is confirmed.

--------------

Another way to verify:

5n+16 = 3(n+8)

5*4+16 = 3(4+8)

20+16 = 3(12)

36 = 36

6 0
2 years ago
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