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2 years ago

Simplify by using the distributive property. -4(x+2)

Mathematics

2 answers:

Vilka [71]2 years ago

6 0

Answer:

-4x - 8

Step-by-step explanation:

-4 ( x + 2 )

( -4 * x ) + ( -4 * 2 )

-4x + ( -8 )

-4x - 8

bagirrra123 [75]2 years ago

4 0

Answer:

-4x-8

Step-by-step explanation:

-4(x+2)

-4(x)=-4x

-4(2)=-8

-4x-8

Hope this helps!

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4x – 3y=8<br> 2. 5x – 2y = -11

AlladinOne [14]

Answer:

$x = -7$

$y = - 12$

Step-by-step explanation:

The given system is:

4x-3y=8

5x-2y=-11

We make x the subject in the top equation to get:

$x = \frac{3}{4}y + 2$

Plug this expression for x in the bottom equation to get

$5( \frac{3}{4}y + 2) - 2y= - 11$

We expand to obtain

$\frac{15}{4} y + 10 - 2y = - 11$

Multiply through by 4

$15y + 40 - 8y = - 44$

Group similar terms

$15y - 8y = - 44 - 40$

$7y = - 84$

$y = - \frac{84}{7} =-12$

This means

$x = -12\times \frac{3}{4} + 2$

$x = - 7$

6 0

3 years ago

Please help, quickly if possible

sleet_krkn [62]

Answer:

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Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Write an equation of the line with the given slope and y intercept: m = 2, b = - 3/4

Alex

Answer:

y = 2x - 3/4

Step-by-step explanation:

Adapt the slope-intercept form: y = mx + b.

Making the indicated substitutions, we get

y = 2x - 3/4

4 0

3 years ago

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What

galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

$P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

$P(D_i)=\frac{1}{6}$ for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

$P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}$

$P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}$

$P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}$

$P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}$

$P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}$

$P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0$

Part A:

The probability that all of the balls selected are white:

$P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)$

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find $P(D_3|A)$

The data required is calculated above:

$P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

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3 years ago

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3 years ago