Recent content by Retro95

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    Stationary Points: (0,0.5) (0,-0.5) (1.25,0) (-1.25,0) Maximum Point

    Thanks for the help I really appreciate it but if you're going to help me then you might as well help me set out the answer since I've done the main working outs Since I've concluded that a stationary point is at (0,0.5) and it's a maximum point since fxx & fyy are negative, also because...
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    Stationary Points: (0,0.5) (0,-0.5) (1.25,0) (-1.25,0) Maximum Point

    Solving both of them simultaneously I got.. 2-10x-4y=0 5-4x-10y=0 Multiplying the top by 4 and the bottom by 10 gives.. 8-40x-4y=0 50-40x-100y=0 The x's cancel out, so.. -42+84y=0 84y=42 y=0.5 Subbing y into the original equation makes x=0.12 I've done the main working...
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    Stationary Points: (0,0.5) (0,-0.5) (1.25,0) (-1.25,0) Maximum Point

    Homework Statement Find the stationary point(s) and also find out whether the stationary point(s) are minimum/maximum z = 2x + 5y - 5y2 - 5x2 - 4xy - 1000 Homework Equations - The Attempt at a Solution I'm not 100% sure if I was correct and laid the answer out properly, can someone please...
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    How Do You Determine Conditional Probabilities in Practical Scenarios?

    Yeah thanks I eventually figured it out after a lot of application question practice. ************************************ Also I have an exam in 2 days but my stupid lecturer decided not to give us the answer sheet to a previous past paper for us to practice and check our answers on. So...
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    How Do You Determine Conditional Probabilities in Practical Scenarios?

    I can easily solve a problem if the probabilities are already defined in the question.. such as P(A|B) But I always get confused about how to actually define the probability from an application exercise. E.g.. Market research shows that (A)75% if customers want text messages on their phone...
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