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2 years ago

Pls dont pwm . will mark brainliest✌

Mathematics

1 answer:

alexgriva [62]2 years ago

5 0

Answer:

Option A

Step-by-step explanation:

We know that the "3x ? 2" line must be positive, so it cannot be C or D. The other line is negative, so the "2x ? 5" line must be -2x. This gets rid of option B, leaving us with our answer of A!

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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Find the solution to the differential equation dx/dy+y/3=0 subject to the initial conditions y(0)=12.

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$\dfrac{\mathrm dx}{\mathrm dy}+\dfrac y3=0\iff\mathrm dx=-\dfrac y3\,\mathrm dy$
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Given that $y(0)=12$ , we have

$0=-\dfrac16(12)^2+C\implies C=24$

So the particular solution to the initial value problem is

$x=-\dfrac16y^2+24\iff y^2=144-6x$

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2 years ago

. Imagine a game of 3 players where exactly one player wins in the end and all players have equal chances of being the winner. T

lbvjy [14]

Answer:

5/9

Step-by-step explanation:

Number of players = 3

number of times game is repeated = 4

P( any person wins a game ) = 1/3

P ( any person does not win a game ) = 1 - 1/3 = 2/3

P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81

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Find the probability that there is at least one person who wins no games

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A1 = player 1 wins no game

A2 = player 2 wins no game

A3 = players 3 wins no game

Applying the inclusion-exclusion formula

P( A1 ∪ A2 U A3 ) = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1 ∩ A3 ) + P( A1 ∩ A2 ∩ A3 )

where

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P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0

Hence

P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9

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