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2 years ago

12

PLS HELP ME PLS PLS !

1 answer:

BaLLatris [955]2 years ago

4 0

Answer:

ill help

Step-by-step explanation:

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Sonoma bikes 5 miles to Paige's house. On a map, they measure that distance as 5/6cm. The same map shows that the mall is 3 1/2c

Elden [556K]

Simplify fraction then subtract fractions

7 0

3 years ago

2 to the power of 5 over minus 3 equals 2 to the power of 5 over minus 3 equals 2 to the power of 5 over minus 3 equals 2 to the

nevsk [136]

You are not asking any question. Maybe you want the way how to write it. If so:

2^5/(-3)=2^5/(-3)=2^5/(-3)

5 0

3 years ago

Can somone answer this

Sunny_sXe [5.5K]

Answer:

$618

Step-by-step explanation:

$1854 divided by 18 months = $103 per month

$103 x 12 months = $1236 paid in one year.

$1854originally owed - $1236 paid in a year = $618 still owed after one year of payments.

8 0

2 years ago

Read 2 more answers

3/x^2-1 - 2(x+3) = 5/x+1

weqwewe [10]

Answer:

X=1.132589

Step-by-step explanation:

If im correct the answer would be x=1.132589

3

x2−1

−2(x+3)=

5

x+1

3

(x+1)(x−1)

+−2x−6=

5

x+1

Multiply all terms by (x+1)(x-1) and cancel:

3+(−2x−6)(x+1)(x−1)=5(x−1)

−2x3−6x2+2x+9=5x−5(Simplify both sides of the equation)

−2x3−6x2+2x+9−(5x−5)=5x−5−(5x−5)(Subtract 5x-5 from both sides)

−2x3−6x2−3x+14=0

(Use cubic formula)

x=1.132589

And dont forget to check your answer

3 0

2 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