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2 years ago

10

Find the probability that when a couple has two children, at least one of them is a boy. (Assume that boys and girls are equal

ly likely.)

1 answer:

Eddi Din [679]2 years ago

3 0

1/2 is awesome answer<span />

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Which expression is equivalent to 2.8k - 8.4

serg [7]

First by simplifying step by step

2.8k-8.4

There are no like terms.

Answer is = 2.8k-8.4

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2 years ago

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2 years ago

Which statement is true? WILL GIVE BRAINLIEST! (If you are right!)

Setler79 [48]

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2 years ago

Read 2 more answers

Please help me with this

Zanzabum

Answer:

x=

−4y+32

9

Step-by-step explanation:

Let's solve for x.

y=

−9

4

x+8

Step 1: Flip the equation.

−9

4

x+8=y

Step 2: Add -8 to both sides.

−9

4

x+8+−8=y+−8

−9

4

x=y−8

Step 3: Divide both sides by (-9)/4.

−9

4

x

−9

4

=

y−8

−9

4

x=

−4y+32

9

I think this is the right answer.

6 0

2 years ago

Solve<br><img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

2 years ago