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1 ounce = 28.3495 grams (perhaps round to 28g)
(use exact figure if more accuracy is needed then round at the end)So for instance how many grams in 6 ounces ?
6x28.3495 = 170.097
How many ounces is 100 grams ?
100 / 28.3495 = 3.527....
To the nearest whole number there are 170 grams in 6 ounces.
To 1 d.p. accuracy there are 3.5 ounces in 100g.
About the email I have just sent:
Just out of interest: I sent that at 12:57 - when did you get the email?
I did wonder whether the length of an email had anything to do with it (eg. 10K rather than 3K for instance).
None of the emails are especially big in today's terms when 10K = 1024 x 10 = 10240 bytes for instance is small compared
to let us say the RAM in a computer let us say 2GB for instance which is 2147483648 (2 x 1024 x 1024 x 1024).
However I remember from something that happened some time ago (maybe a year or 2) where I had to send a large attachment
(50+MB approx ???) that someone had asked for where my email system warned me that it couldn't send it and I received a delivery
failure message. To give an idea of scale this is about 5.24 million bytes more than 5000 times the size of a 10K email.
You should really get an error report or delivery failure message occur every time something you send does not get through.
However maybe some software is set up not to do this sometimes, or maybe there is some other problem.
My experience today and recently is that all of your emails are getting through as far as I know. You have sometimes spelled
my email address wrong, but your computer will put the address in for you in a reply and may help you even when sending a new
message depending on what you are using to send them with and how good the software is.
I have thought of another conversion you might like to do:
ounces to grams and grams to ounces
eg. 1 ounce = 28.3495 grams (perhaps round to 28g)
(use exact figure if more accuracy is needed then round at the end)
So for instance how many grams in 6 ounces ?
How many ounces is 100 grams ?
Be careful to make sure you have got that right if needed for a cooking
recipe and/or something important.
(2) Convert 20 inches into centimetres
I am getting:
20 x 2.54 = 50.8 cm
It may be helpful to think of a ruler (eg. 30cm, 15cm etc.)
to get a rough idea when doing these to make sure you don't
do the calculation the wrong way round.
I don't think the weather has had any effect on the system (overheating etc.).
The thing that I thought was the weather turned out to be caused
by roadworkers making an error causing a connection failure.
I received the 2 emails where you asked for the questions only too
late and had just sent the answers and questions emails.
After a pause of 5 minutes or so I sent the questions only.
It may still help you to try to do those since you have the answers now
in case you are stuck with them. Notice also that I have answered two
of the recent maths exercises as examples in case you were stuck on them
(see above).
I have received some emails from you today.
I have just replied with questions and answers from the 17th May and
the 21st April, which were sent as answers on 20/21 June.
They are long emails so if needed I can post here, but if you now have them
then great.
A few on metres, feet, inches and centimetres:
(1) Convert 5 metres into inches.
I am also going to do this one:
5 metres = 500 cm
To convert to inches I need to divide by 2.54
500 / 2.54 = 196.85 inches (to 2 dp.)
or 197 to the nearest inch.
One thing you might like to consider is:
An object is said to weight one stone exactly.
1 stone = 14 pounds
1 kg = 2.20462 poundsHow much is one stone in kg ?
By kg or Kg or KG I mean kilograms that is to say 1000 grams.
"kilo" usually means 1000 of something.
I am going to do this as an example:
1 stone is 14 pounds
Therefore we want 14 pounds converted into kg.
If you divide 2.20462 pounds by 2.20462 you get 1 kg.
So divide 14 pounds by 2.20462 using a calculator and get:
6.350300732
Obviously this needs to be rounded because it is more accurate than we could
possibly have measured so I am going to round it to 6.35 kg in one stone.
(I have found out from another source that it is 6.35029 - I am not sure which is more
accurate and it does not matter because in practice you would not need to be that
precise except in engineering/scientific precision situations.)
I agree with bobbym in that it is the second line where the problem starts.
The square root function can only be used in this way if a disabiguation is made where
we either take the positive value so reverse a square of a positive number, or I suppose
take the negative value provided the original number was negative before being squared.
If there is no guarantee of the sign then both negative and positive cases of the number
squared then square rooted should be checked to see whether the equation is still true
in every case. Otherwise a true statement can imply an untrue one as you did purposefully.
However in the quadratic formula for instance the square root sign is used to indicate
that you can take both the positive and negative square root so the square root has to
allow you to take either the positive or negative case.
So let us start with a negative number so -1 then square it and get 1 then square root it
and we get 1. That is the trouble with using a function like square root as an inverse.
It is good for calculating things, but in a formal proof I find it difficult, because care has
to be taken for instance not to accidentally make a sign error which I have sometimes
done by accident in a tutor marked assignment (I remember I accidentally fudged the sign
of an otherwise correct proof and had a couple of marks taken off. I think that I had done
the proof first with one sign error which I knew must be there somewhere. I searched for
the "error" and "corrected it" only to find that according to the tutor it had 2 sign errors
which cancelled).
Anyway your example gives a good case something of the kind happening with a very
amusing consequence.
I think the answer might be: (0.099 / (0.089 + 0.099)
Reason:
Let us assume that P(spam) = 0.1
and that P(not spam) = 0.9
Now I will consider the two possibilities of flag and not flag as if it were an event afterwards:
Spam Event Flag Event Probability of Spam Event AND Flag Event
True(0.1) True (0.89) 0.089 *
True(0.1) False(0.11) 0.011
False(0.9) True (0.11) 0.099 *
False(0.9) False(0.89) 0.801
The two that I have put * next two are very significant in this because both concern a
situation where the flag event is true. So in these cases the message has been flagged
as spam. So the total probability of this is (0.089 + 0.099).
We are also interested in finding the probability of the situation where the spam event
was false but the flag event was true individually. This is (0.099).
Hence P(not spam | flag ) = 0.099 / (0.099 + 0.089) = 0.526596 (to 6 dp.)
If you do that style of analysis with the assumption that P(flag) = 0.1
then it gives a rather silly answer of 0.11 - silly because it only uses half the tree
and does not use Bayes or any conditional logic and is just the probability of an incorrect
flagging of something not a spam under this assumption on the basis of (1-0.89) = 0.11
This seems too simple to be true for this type of problem.
So I prefer the answer 52.65 .... etc
(0.10*0.11 + 0.9*0.89)
To my mind this reads as P(flag | not spam) x P(flag) + P(~flag) x p(~flag | not spam)
Which using Bayes simplifies to: P (flag and not spam) + P (~flag and not spam)
Which perhaps also simplifies to: P (not spam)
0.10*0.11 / (0.10*0.11 + 0.9*0.89)
This I thought came from: P(flag and not spam) / P(not spam)
Is that what you meant and why you made that calculation?
If you let A = flag
and let B = not spam
Then apply Bayes of P (A and B) = P (A | B ) x P (B)
then this supports this logic.
But P(A | B) = P (flag | not spam)
unless I am mistaken we want P (not spam | flag)
So we want P (B | A) perhaps ?
P (B | A) x P(A) = P(B and A)
The trouble with that argument is that it does not work. Perhaps your answer is correct.
because P(B and A) = P (A and B)
so P (B | A) = P(A and B) / P(A)
The problem with that is that it gives us simply the answer 0.11 which does not make sense.
It would be daft to ask that question so are you sure you have formulated everything correctly
and consistently.
Make sure you use the term "flag" consistently and do not confuse with "spam".
Is 0.1 definately the probability that a random message is spam?
Not 0.1 being the probability that a random message is flag?
given that 1/10 of message is a spam message.
Does this mean that 0.1 of all messages are flagged as spam ?
Or that 0.1 of all messages are spam ?
So P(flag) = 0.1
Or is it P(spam) = 0.1 ? (I think this version is more likely, but you have used the other)
I will have to stop for now. I am sharing your confusion about this it is a complicated problem
and the formulation of the probabilities originally is not very clear.
I am wondering whether it might help to draw a diagram similar to this adapting where needed
to suit the problem:
Spam event Flag event
Spam - true -true spam and flag
-false spam and not flag
- false -true not spam AND flag
-false not spam AND not flag
In theory Bayes theorem can calculate P (A and B) = P(A | B) x P(B)
The trouble is do we really know the values to plug in to that formula ?
The detector has just detect a message as being spam. what is the probability that the message is actually not a spam message?
So we are looking for: P(not spam given that the message is flagged as spam)
Or using your notation: P(~spam | flag)
I am a little confused myself and I may need to think some more about this.....
I am not sure either way at the moment I am trying to draw a tree diagram on paper to help ....
You want:
P (not spam | flag)
P (not spam | flag) = P (not spam and flag) / P (flag)
How do we work out P (not spam and flag) ?
given that 1/10 of message is a spam message.
Does this mean that 0.1 of all messages are flagged as spam ?
Or that 0.1 of all messages are spam ?
So P(flag) = 0.1
Or is it P(spam) = 0.1
spam detector will correct identify a message as spam 89 percent of the time.
the spam detector will correct identify a message as not spam 89 percent of the time too.
Another question about feet, inches, cm and metres is:
A person is said to be 5 foot 9 inches tall exactly.
(1) How many inches is the person is just using inches ? (Hint: 12 inches equals one foot.)
(2) Now convert this into centimetres. (Remember: 1 inch = 2.54 cm)
(3) Now convert this into metres.
(4) You could also put this in the form of metres and centimetres.
(I mean by that to put 100cm as 1 metre and give the rest as centimetres.)
A few on metres, feet, inches and centimetres:
(1) Convert 5 metres into inches.
(2) Convert 20 inches into centimetres
(3) Convert 15cm into inches.
(4) Convert 1 foot into metres
One on weight:
(5) Convert 1 pound into kg (use the information in the previous post to help with this)
Doing Q5 might help with the question:
(6) Convert 1 stone into kg.
One thing you might like to consider is:
An object is said to weight one stone exactly.
1 stone = 14 pounds
1 kg = 2.20462 pounds
How much is one stone in kg ?
By kg or Kg or KG I mean kilograms that is to say 1000 grams.
"kilo" usually means 1000 of something.
I am getting 3.28 feet in a metre to 2 decimal places.
Not sure if this helps, but imagine a 1 metre long ruler - now imagine 3 foot long (12in or 30cm)
rulers in a line. The metre ruler is a little bit longer than the 3 one foot long rulers.
No 100 / 2.54 = 39.37 inches approx.
Notice that 39.37.. x 2.54 = 100 cm
You can if you like divide by 12 to get the amount of feet in one metre.
Correct. 30.48cm is 12 inches.
Right are you able to work out how many feet there are in one metre ?
Or if it is easier how many inches are there in 100cm ? (Then how many feet in 100cm?)
All correct well done (except Q3). Try the one above if you haven't already.
Actually Q3 was 5000 /100 = 50 m
Anyway you have got the hang of it. Good.
I will go away for a few minutes for you to try the feet/inches/cm one.
Hi Mandy
I have decided to give you just ONE question which is one I have already mentioned:
1 foot = 12 inches
1 inch = 2.54 cm
How many cm are there in a foot, preferably to 2 decimal places?
Do not spend too long over that. If you cannot answer within 5 minutes just guess the answer.
You can use a calculator for that one, but tell me the method that you used.
I have been sent 3 emails to the BT address. Two were about 5:11pm one was about 9.43am.
I have replied to one of them. The two at about 5:11pm were "Blackberry" ones.
The one at about 9.43am was a mobile one.
Do you want me to give you the answer with a brief method for one or two of the recent questions?
Another idea I have had is for me to meet you online to ask just one question at a time for you
to try and me to give "correct" or "wrong" and say why if it is wrong and what it should be?
This has worked well before with some maths problems that you were stuck on.
Hi Mandy. I have replied to the gmail now.