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3 years ago

15

2/3(2x+1) - 1/2(3x-2)=2 solve for x

1 answer:

bagirrra123 [75]3 years ago

3 0

Answer:

x= -2

Step-by-step explanation:

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If it takes 6 hours for a plane to travel 720km with a tail wind and 8 hours to make the return trip with a head wind. Find the

denis-greek [22]

The speed of wind and plane are 105 kmph and 15 kmph respectively.

<u>Solution:</u>

Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.

We have to find the air speed of the plane and speed of the wind.

Now, let the speed wind be "a" and speed of aeroplane be "b"

And, we know that, distance = speed x time.

$\text { Now, at tail wind } \rightarrow 720=(a+b) \times 6 \rightarrow a+b=\frac{720}{6} \rightarrow a+b=120 \rightarrow(1)$

Now at head wind → $720=(a-b) \times 8 \rightarrow a-b=\frac{720}{8} \rightarrow a-b=90 \rightarrow(2)$

So, solve (1) and (2) by addition

2a = 210

a = 105

substitute a value in (1) ⇒ 105 + b = 120

⇒ b = 120 – 105 ⇒ b = 15.

Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.

Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.

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3 years ago

How can i find values if h(-4) h(x)=1/2x-1???

anzhelika [568]

plug in -4 in for x. then solve.

h(-4)=1/2(-4)-1

h(-4)= -3

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3 years ago

Multiple choice math question.

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3 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

3 years ago