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    Doing the new spec

    I cant seem to figure out how i got question b wrong...
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    There are three scenarios where you lose at least once.

    win-lose, lose-win, lose-lose

    You have only calculated the probability of win-lose.
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    For b), you have to consider every possible case where she loses a match, since she loses at least one.

    This means that the possibilities you have to consider are, if you say that if she wins, the event is W, and if she loses, the event is L...

    W L
    L W
    and L L,

    where W has a probability of 0.7 and L has a probability of 0.3.

    Since all of these outcomes are separate, you have to add their probabilities together.

    The associated probabilities are (0.7 x 0.3) (WL), (0.3 x 0.7) (LW) and (0.3 x 0.3) (LL.)

    You can confirm you have the right answer because in this case there are only four probabilities. If you take your answer away from 1, you should get your answer to a) since you have removed three probabilities by doing so. (Also, you could have answered this by doing 1 - answer from a) at the start, but that won't be so helpful in any case where there are more than two outcomes of an event or in more difficult case.)

    Hope this helps, and feel free to ask if you need any clarification.
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    (Original post by Astro_Joe)
    For b), you have to consider every possible case where she loses a match, since she loses at least one.

    This means that the possibilities you have to consider are, if you say that if she wins, the event is W, and if she loses, the event is L...

    W L
    L W
    and L L,

    where W has a probability of 0.7 and L has a probability of 0.3.

    Since all of these outcomes are separate, you have to add their probabilities together.

    The associated probabilities are (0.7 x 0.3) (WL), (0.3 x 0.7) (LW) and (0.3 x 0.3) (LL.)

    You can confirm you have the right answer because in this case there are only four probabilities. If you take your answer away from 1, you should get your answer to a) since you have removed three probabilities by doing so. (Also, you could have answered this by doing 1 - answer from a) at the start, but that won't be so helpful in any case where there are more than two outcomes of an event or in more difficult case.)

    Hope this helps, and feel free to ask if you need any clarification.
    Thank you so much
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    How do i solve CB?
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    (Original post by z_o_e)
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    Hi Zoe. Could you remember to rotate your photos before attaching them please?
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    (Original post by z_o_e)
    How do i solve CB?
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    You could use trigonometry to find angle B and then use it again to find the length required.
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    (Original post by Mr M)
    You could use trigonometry to find angle B and then use it again to find the length required.
    I don't understand i need to find CB :/
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    (Original post by z_o_e)
    ...
    If you had the angle at B, can you see how you would calculate length CB using this angle and the given lengths (in particular, the length AB)?
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    (Original post by SherlockHolmes)
    If you had the angle at B, can you see how you would calculate length CB using this angle and the given lengths (in particular, the length AB)?
    Yus I solved it by trigonometry and got 67 degrees.
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    (Original post by z_o_e)
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    Good. You can now calculate the length CB using trigonometry for the triangle ABC. Also, leave your angle as 67.38... rather than rounding.
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    (Original post by SherlockHolmes)
    Good. You can now calculate the length CB using trigonometry for the triangle ABC. Also, leave your angle as 67.38... rather than rounding.
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    Once you've got that angle look at the big triangle and I believe from your diagram its a right angled triangle? So use trig again to find the hypotenuse. You have an angle and the adjacent length and you want to find the hypotenuse. What do you do next?
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    (Original post by lordyP)
    Once you've got that angle look at the big triangle and I believe from your diagram its a right angled triangle? So use trig again to find the hypotenuse. You have an angle and the adjacent length and you want to find the hypotenuse. What do you do next?
    I legit give up.

    I cant find the hypotenuse://


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    (Original post by lordyP)
    Once you've got that angle look at the big triangle and I believe from your diagram its a right angled triangle? So use trig again to find the hypotenuse. You have an angle and the adjacent length and you want to find the hypotenuse. What do you do next?
    To find the other triangle i have the length Name:  ImageUploadedByStudent Room1483730995.882958.jpg
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    This is ABC Name:  ImageUploadedByStudent Room1483731076.190202.jpg
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    So Cos(x) = adj/hyp ( just sohcatoa )
    Hyp = adj/cos(x) (rearrange)
    hyp = 13/cos(67.38) (sub in values)
    hyp = 33.8
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    Don't round off your vales too early in the calculations. Leave all values to at least 2dp until the very final answer and even then write it to 2+dp and then round it
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    Help on this please


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    (Original post by z_o_e)
    Help on this please


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    Work out the area of the cross section first before multiplying it by 20 which is the depth of the prism.
 
 
 
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