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2 years ago

6

Solve for b in the literal equation y = 12x + 12b. b=

2 answers:

pantera1 [17]2 years ago

6 0

Answer:

$b = \frac{y-12x}{12}$

Step-by-step explanation:

subtract 12x from both sides, then divide by twelve.

Reptile [31]2 years ago

6 0

Subtract 12x from both sides then divide by 12

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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

2 years ago

A bicyclist travelled for T = 0.7 hours at a constant speed of R = 14 kilometers (km) per hour. Use the formula D = R · T to fin

jeka94

Answer:

9.8

Step-by-step explanation:

We have the following formula:

$D = R \times T$

We have that:

$R = 14, T = 0.7$

So

$D = 14*0.7 = 9.8$

The total distance traveled in kilometers is 9.8

5 0

2 years ago

Read 2 more answers

if a pair of standard dice are rolled what is the probability that one die shows a six but not both of them. reduced fraction an

Novay_Z [31]

Answer:

$P=\frac{5}{18}$

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

so

Let

x------> size of the event space

y-----> size of the sample space

if a die is rolled the sample space is

$\{1,2,3,4,5,6\}$

if two dice rolled the sample space is

$\{1,2,3,4,5,6\}*\{1,2,3,4,5,6\}$

so

The possible pairs are

$y=6*6=36$

The size of the event space x is

$\{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$x=10$

$P=\frac{x}{y}$

substitute

$P=\frac{10}{36}$

Simplify

$P=\frac{5}{18}$

6 0

2 years ago

Which expression is equivalent to the given expression? 2x^2-11x-6

Sveta_85 [38]

Answer:

X=-1/2,6

Step-by-step explanation:

3 0

3 years ago

What is 65% out of 45?

masha68 [24]

Answer:

29.25 is ur answer

Step-by-step explanation:

8 0

2 years ago