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elena55 [62]
3 years ago
14

Alfred draws candles randomly from a pack containing 4 colored candles of the same shape and size. There are 2 red candles, 1 gr

een candle, and 1 blue candle. He draws 1 candle and then draws another candle without replacing the first one. Find the probability of picking 1 red candle followed by another red candle, and show the equation used.
Mathematics
1 answer:
bekas [8.4K]3 years ago
3 0
4 total. And it says "WITHOUT REPLACING" so, 
it is 2/4 times 1/3 = 1/6
So 1/6 is your answer!
GOOD LUCK!! c:
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Name The constant in the expression <br> 2X^4+3X^3-5X^2-8X+12
Vitek1552 [10]

the constant term is 12

All the other terms have variables attached to them

6 0
3 years ago
The equation y – 9 + 9 = –17 + 9 is an example of which property of equality?
frutty [35]
You add 9 to both sides to get:
<span>y – 9 + 9 = –17 + 9
</span>
so it's <span>Addition Property of Equality

answer

</span>B.Addition Property of Equality
4 0
3 years ago
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Solve for x in this equation.
arsen [322]

Answer:

33

Step-by-step explanation:

1. (x-10)+67=90 because it's a 90 degree angle

2.subtract 67: x-10=23

3. add 10: x=33

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3 years ago
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The line y =3x-5 meet x-axis at the point M. The line 3y+2x=2 meets y-axis at point N. Find the equation of the line joining M a
Law Incorporation [45]

<u>Answer-</u>

<em>The line equation is,</em>

\boxed{\boxed{6x+15y-10=0}}

<u>Solution-</u>

The line y =3x-5 meets x-axis at the point M, i.e M is the x-intercept of this line. At the x-intercept y=0, so

\Rightarrow 0 =3x-5

\Rightarrow 3x=5

\Rightarrow x=\dfrac{5}{3}

So, coordinate of M is (\dfrac{5}{3},\ 0)

The line 3y+2x=2 meets y-axis at point N, i.e N is the y-intercept of this line. At the y-intercept x=0, so

\Rightarrow 3y+2(0)=2

\Rightarrow 3y=2

\Rightarrow y=\dfrac{2}{3}

So, coordinate of N is (0,\ \dfrac{2}{3})

The line joining M and N can be found out by applying two point formula of straight line,

\Rightarrow \dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}

\Rightarrow \dfrac{y-0}{\frac{2}{3}-0}=\dfrac{x-\frac{5}{3}}{0-\frac{5}{3}}

\Rightarrow \dfrac{y}{\frac{2}{3}}=\dfrac{x-\frac{5}{3}}{-\frac{5}{3}}

\Rightarrow -\dfrac{5}{3}y=\dfrac{2}{3}(x-\frac{5}{3})

\Rightarrow -5y=2(x-\dfrac{5}{3})

\Rightarrow -5y=2x-\dfrac{10}{3}

\Rightarrow 2x+5y-\dfrac{10}{3}=0

As it is given that all the coefficients are integers, so multiplying with 3

\Rightarrow 6x+15y-10=0

4 0
3 years ago
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Answer this for me plz :))
Naily [24]
Mmmm c i’m thinking it’s that
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2 years ago
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