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What's your question? You need EFA?

My question is how do you calculate EFA?
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What's your question? You need EFA?
I found it already...thank you missy!question 13 is solved in Alevels thread. Rizwan solved it. I'll tag you there.
If you remove the lines CB and FE you will get a regular parallelogram. Diagonal inside a parallelogram will bisect the angle and angle EDA will become 70. And so the same diagonal will bisect FAB and make 70. After calculating these angles draw back the lines you omitted EF and CB.Therefore, 360 - 70 - 70 - 120My question is how do you calculate EFA?
OhhhI found it already...thank you missy!![]()
Got it...Thank youIf you remove the lines CB and FE you will get a regular parallelogram. Diagonal inside a parallelogram will bisect the angle and angle EDA will become 70. And so the same diagonal will bisect FAB and make 70. After calculating these angles draw back the lines you omitted EF and CB.Therefore, 360 - 70 - 70 - 120
100.
I hope u get it.
happiness is that I don't need that anymore! only taking AS this session!Ohhh![]()
unable to be read![]()
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Can someone please help with the c part.
TIA.![]()
sorry ... will post a clearer picunable to be read![]()
b ii - 56/165?Unable to solve last two parts , can someone please explain it ? (Step by step)
Yeah .. Same answersb ii - 56/165?
iii - 1/5?
^ are these the answers? I'll post the solutions if these will be correct.
The first part says that there shouldn't be any E in three selections of Card.Unable to solve last two parts , can someone please explain it ? (Step by step)
Alright , can you solve one more question too ? (Last part ) thanksThe first part says that there shouldn't be any E in three selections of Card.
There are three Cards with Letter E. So the probability of drawing a non-E card ( let it be E' ) for the first time is 8/11.
For the second when he'll pick a card, the probability of drawing E' is 7/10. The sample space has reduced to 10 because the first card picked up has not been replaced.
Third time, the probability of picking E' will be 6/9.
So the probability that no Card with letter E has been picked up in three tries is = 8/11 * 7/10 * 6/9 = 56/165
ii - You can see that there are only two letters that repeat in the word given. 3Es and 2Ns.
First consider that the three cards picked up contain 2Es and one non-E (E'). This can be done in 3 different ways:
E E E' OR E E' E OR E' E E
so the probability that 2Es are there in the three selections is : 3 * 3/11 * 2/10 * 8/9
Now consider that the selections contain 2 Ns and one Non-N (N'). This can also be done in 3 different ways:
N N N' Or N N' N Or N' N N
so the probability that 2Ns are there in the selections is : 3 * 2/11 * 1/10 * 9/9
So the combined probability will be : 3 * 3/11 * 2/10 * 8/9 + 3 * 2/11 * 1/10 * 9/9 = 1/5
OR
more simple it can done in this way: ( 2C2 * 9C1 + 3C2 * 8C1 ) / 11C3 = 1/5
Alright , can you solve one more question too ? (Last part ) thanks
Second oneWhich part? There're two files, so which one?
Sorry for getting late :/ I went away for a very important task.Second one
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