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3 years ago

Can someone help me with the equation in the picture below.

Mathematics

1 answer:

zlopas [31]3 years ago

4 0

Answer:

x=3

Step-by-step explanation:

First, you would distribute the 4 to the (x-3), giving you 2x-6+4x-12

From there combine like terms, 2x+4x=6x, -6+-12=-18

6x+-18=0

6x=18

divide both side by 6, and you get 3=x!

I hope that all makes sense :)

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There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

3 years ago

Brian can type 1 and one-third pages every 20 minutes. How many pages can he type in 2.5 hours? 7.5 pages 10 pages 22.5 pages 20

vladimir2022 [97]

Answer:

10 pages

Step-by-step explanation:

1 and 1/3 pages multiplies by 7.5 brings us to the answer of ten pages

20 minutes goes into 2 and 1/2 hours 7.5 times

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4 years ago

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kirza4 [7]

Answer:

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4 years ago

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Aliun [14]

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Step-by-step explanation:

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3 years ago

How do how you cross cancel multiplying fractions

DENIUS [597]

It's easy once you spot the ones that can cross cancel!

Say we have the fractions 8/10 and 20/23. $\frac{8}{10} \frac{20}{23}$ (it's easier to see on top of each other)

If you look diagonally , so 8 and 23 and 10 and 20, you can see that 10 and 20 have a common factor. So we divide it by the highest common factor to reduce those numbers, making it easier to multiply. 10 and 20 can become 1 and 2, dividing by 10. So now we are left with 8/1 and 2/23, and now we multiply normally going across so 16/23.

This works going both diagonals and simplifying both, but in that case it would be easier to try and simplify the fractions before cross multiplying them.

Basically: look for those diagonals and if they can be divided down by the highest common factor, go for it to make it easier to multiply normally afterwards.

Hope I helped!

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3 years ago