Ask question

Ask question

All categories

2 years ago

8

Solve the system. y = 3x + 1 -x + 2y = -3

1 answer:

PIT_PIT [208]2 years ago

3 0

-3, -3

,........... .......

You might be interested in

Each day Diego's mother drives to her office in Baltimore. The driving distance from home to Baltimore, Maryland is 7 miles. On

Sidana [21]

Answer:

D

Step-by-step explanation:

(7+8)*3

6 0

2 years ago

Dan makes a pineapple sorbet. He uses 3 lemons. Work out how many people he makes pineapple sorbet for.

seropon [69]

Answer:

N/A

Step-by-step explanation:

There is not enough information to answer with a number.

6 0

2 years ago

A spheres radius is multiplied by 2. How is it’s surface area affected

dexar [7]

Answer:

The surface area is multiplied by 3.

Step-by-step explanation:

If r is the radius of a sphere its surface area = 4(pi)r^2. If the radius is doubled then its surface area = 4(pi)(2r)^2 = 4*4(pi)r^2. So the surface area becomes 4-fold or the increase in surface area is 300%.

3 0

3 years ago

Covert 4.12 in a faction

postnew [5]

Answer:

it would be 103/25

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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