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3 years ago

Help please I really need it

Mathematics

1 answer:

gtnhenbr [62]3 years ago

7 0

So we want to solve 2x − 40 = 3y for x2x − 40 + 40 = 3y + 40(Add 40 to both sides!)2x = 3y + 40
Now divide both sides by 2!
 $\frac{2x}{2} = \frac{3y + 40}{2} \ \textgreater \ x = \frac{3}{2}y + 20$
Now substitute $\frac{3}{2}y + 20 for x in 2y = -4x + 48

$
$2y = -4( \frac{3}{2}y + 20) + 48$
Which then gives us..
2y = −6y − 32(Make sure to simplify both sides of the equation)2y + 6y = −6y − 32 + 6y(Now add 6y to both sides)8y = −32
Now divide both sides by 8
 $\frac{8y}{8} = \frac{-32}{8} \ \textgreater \ y = -4$
Now we have to substitute -4 for y in $x = \frac{3}{2}y + 20 \ \textgreater \ \frac{3}{2}(-4) + 20 \ \textgreater \ x = 14$
So we end up getting...
x = 14 and y = −4.
Hope this helps!

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The question is incomplete:

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$P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\$

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