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3 years ago

12

Pls help!!! What situation can this graph represent?

1 answer:

xxMikexx [17]3 years ago

3 0

Answer: Ocean tides

Step-by-step explanation: when the moon orbits the earth, it's gravitational pull causes tidal movements in large bodies of water; such as the oceans.

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Which Expression is equivalent to 48z +42q<br><br> A) 6(8z+7q)<br><br> B) 6*8z+7q<br><br> C) 8z+7q

Butoxors [25]

Answer:

6(8z+7q)

Step-by-step explanation:

48z+42q

Factor out 6.

6(8z+7q)

5 0

2 years ago

The record high temperature in Alabama is 112°F and the record low temperature is −27°F. What is the difference between the high

Minchanka [31]

The difference between the highest and lowest temperature is 85°F.

7 0

3 years ago

Read 2 more answers

Please solve please and thank you

Dahasolnce [82]

(2x+2)(3x+1)
2x*3x+2x*1+2*3x+2*1
6x^2+2x+6x+2
6x^2+8x+2

Hope this helped

6 0

3 years ago

Read 2 more answers

What are all the power of threes through 1000

algol13

1000= 10^3 -- ten to the third power

assuming you mean whole numbers,

graba calculator, figure out what
1^3, 2^3 3^3 ... 9^3 are
or
1x1x1
2x2x2
3x3x3
...
9x9x9

6 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

3 years ago