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3 years ago

8

What is the value of x?

2 answers:

mote1985 [20]3 years ago

7 0

Answer:

Download photoath it really helps

8090 [49]3 years ago

3 0

Answer:

62 degrees

Step-by-step explanation:

X and 28 degrees are on a 90 degree corner, 28 degrees needs to be subtracted from 90 degrees to get the value of x so, 90 - 28 = 62 degrees

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How to make a circeldiagramm

lys-0071 [83]

Make a circle and then another circle. There should be a oval in the middle.
Like this:
https://docs.google.com/drawings/d/10jIdkQW5rnxfDS9T2WeqkwRTvl92oJsYkXVLj8Zn6hU/edit

4 0

2 years ago

Jack spent 1/3 hour cleaning his room which fraction is equivalent

Vinil7 [7]

Answer:

2/6

Step-by-step explanation:

1/3

Write an equivalent fraction by multiplying

1/3 * 2/2

2/6

I hope this was useful!

4 0

3 years ago

A box containing packets of oatmeal has the shape of a rectangular prism measuring 12 cm by 20 cm by 7 cm.

qaws [65]

The answer is D

12 * 20 * 7 = 1680

8 0

3 years ago

For what value of x does 3^4x=27^x-3? a. –9 b. –3 c. 3 d. 9

Irina18 [472]

$\bf 3^{4x}=27^{x-3}\implies 3^{4x}=(3^3)^{x-3}\stackrel{~\hfill \textit{same base, the exponents must also be the same}}{\implies 3^{4x}=3^{3(x-3)}\implies 3^{4x}=3^{3x-9}\qquad } \\\\\\ 4x=3x-9\implies x=-9$

5 0

3 years ago

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

2 years ago