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3 years ago

7

Can somebody please help me solve this algebra problem: 3p+2r=n solve for p

1 answer:

Lera25 [3.4K]3 years ago

8 0

3p + 2r = n.....subtract 2r from both sides
3p = -2r + n...divide both sides by 3
p = (-2r + n)/3 or p = -2/3r + n/3 or p = -2/3r + 1/3n

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How many degrees are in the following sum?

Amiraneli [1.4K]

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3 years ago

What is the simplified form of (-3x3y2) (5xy-1)

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3 years ago

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

valina [46]

Answer:

$P(Same)=\frac{61}{190}$

Step-by-step explanation:

Given

$Red = 5$

$White = 6$

$Black = 9$

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

$Total = 5 + 6 + 9$

$Total = 20$

This is calculated as:

$P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)$

So, we have:

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

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

$P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1} + \frac{9}{20} * \frac{9- 1}{20- 1}$

$P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19} + \frac{9}{20} * \frac{8}{19}$

$P(Same)=\frac{20}{380} + \frac{30}{380} + \frac{72}{380}$

$P(Same)=\frac{20+30+72}{380}$

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4 0

2 years ago

Can someone help me with this question please.

Daniel [21]

Answer:

c

Step-by-step explanation:

*View attached graph*

7 0

2 years ago

OC and OR are similar. Find the length of QP.

Elena L [17]

Given:

Circle C and circle R are similar.

The length of arc AB is $s = \frac{22 \pi}{9}$

The radius of circle C (AC) = 4 unit

The radius of circle R (QR) =6 unit

To find the length of arc QP.

Formula

The relation between s, r and $\theta$ is

$arclength = 2\pi r \frac{\theta}{360}$

where,

s be the length of the arc

r be the radius

$\theta$ be the angle.

Now,

For circle C

Taking r = 4

According to the problem,

$2 \pi r \frac{\theta}{360} = \frac{22 \pi}{9}$

or, $2r \frac{\theta}{360} = \frac{22}{9}$ [ eliminating $\theta$ from both side]

or, $\theta = \frac{(22)(360)}{(9)(2)(4)}$

or, $\theta = 110^\circ$

Again,

For circle R

Taking, r = 6 and $\theta = 110^\circ$ we get,

The length of arc QP is

$arc length = 2\pi (6)(\frac{110}{360} )$

or, $arclength = \frac{11 \pi}{3}$

Hence,

The length of QP is $\frac{11 \pi}{3}$ . Option C.

5 0

3 years ago