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3 years ago

Help me with this plsss

Mathematics

1 answer:

slavikrds [6]3 years ago

6 0

Answer:

56.31 is r

Hope it helps!!!

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Find the sum of 5x+20 and the opposite of 20. Write an equivalent expression in standard form.

baherus [9]

5x+(-20)=-15x and there you have it

5 0

4 years ago

PLEASE HELP ILL MARK BRAILIEST!!! P.S please show how you figured it out

Arturiano [62]

Answer:

I think that the answer is either B or C

Step-by-step explanation:

I concluded to this answer simply by looking at where the lines were placed.

Remember that if a line is not touching, but next to another line, they are parallel to each other? Well, that doesn't change here. since, {m angle 1} and {m angle 2} added together equal 180, I would believe that the answer is C, but I may be wrong.

Well, I hope that helps :)

5 0

3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

We used Total - 1 because it is a probability without replacement

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$