Maths, Addmaths and Statistics: Post your doubts here!

[Minato112]

Check here and tell me: http://www.s-cool.co.uk/gcse/maths/transformations

 

[zainabkhann]

Check here and tell me: http://www.s-cool.co.uk/gcse/maths/transformations

This is exactly what i was looking for ! Thankyou, You have been a great help. Could you help me on question 6 from paper 1 . I need a thorough explanation as to what we are suppose to do

 

[Minato112]

This is exactly what i was looking for ! Thankyou, You have been a great help. Could you help me on question 6 from paper 1 . I need a thorough explanation as to what we are suppose to do

Area of shaded region
= Area of square - Area of circle
= (8 * 8) - ( ("pie")((3)^2)
= 64 - 9"pie"

 

[parthrocks]

Area of shaded region
= Area of square - Area of circle
= (8 * 8) - ( ("pie")((3)^2)
= 64 - 9"pie"

Ya thats what i got

 

[zainabkhann]

Area of shaded region
= Area of square - Area of circle
= (8 * 8) - ( ("pie")((3)^2)
= 64 - 9"pie"

How do we get 9? its piexr^2 so 3.14x3^2 is 28

 

[zainabkhann]

How are u guys getting 9 as the area of the circle? :/

 

[Minato112]

How are u guys getting 9 as the area of the circle? :/

Leave the pie as it is... and 3^2= 9.. U understand now?

 

[zainabkhann]

Leave the pie as it is... and 3^2= 9.. U understand now?

Yeah i do thanks But i have a question why do we leave the pie?

 

[scouserlfc]

Yeah i do thanks But i have a question why do we leave the pie?

When u solve it actually u dont leave the pie u solve it its pretty much obvious that minato112 had no calculator at his disposal while he was writing

 

[zainabkhann]

When u solve it actually u dont leave the pie u solve it its pretty much obvious that minato112 had no calculator at his disposal while he was writing

could you show me what you mean on writing like type the solution here and show me please thanks

 

[Minato112]

When u solve it actually u dont leave the pie u solve it its pretty much obvious that minato112 had no calculator at his disposal while he was writing

could you show me what you mean on writing like type the solution here and show me please thanks

The question specified to leave the answer in terms of pie.. Thats Y...

 

[zainabkhann]

The question specified to leave the answer in terms of pie.. Thats Y...

Oh okey thanks then, could you check out question 8 from the same past paper, i am getting the upper bound to be 1.4 whereas in the marking scheme it says its : 9.5. Need help in both parts of the question. thanks

 

[zainabkhann]

Question 23 guys b and c
and please give an explanation as to why that is the answer.

 

[Minato112]

Oh okey thanks then, could you check out question 8 from the same past paper, i am getting the upper bound to be 1.4 whereas in the marking scheme it says its : 9.5. Need help in both parts of the question. thanks

0.9m = 90 cm
Now to the nearest ten centimeters

10 cm/2 = 5 cm

So the upper bound is 90cm + 5 cm = 95 cm = 0.95 cm (Btw u sure the mark scheme says 9.5?)

 

[Minato112]

Question 23 guys b and c
and please give an explanation as to why that is the answer.

(b) The modal class : It is the one having the highest frequency...

Well i didnt draw the graph but i fink this information is enough. Just look the time which have the highest frequency...

(c) U only have to add the frequencies of the times above eight, that is, 7 + 4 + 2 = 13

 

[zainabkhann]

0.9m = 90 cm
Now to the nearest ten centimeters

10 cm/2 = 5 cm

So the upper bound is 90cm + 5 cm = 95 cm = 0.95 cm (Btw u sure the mark scheme says 9.5?)

Sorry it says 0.95m

 

[Minato112]

Sorry it says 0.95m

Ok I solved it then...

 

[zainabkhann]

(b) The modal class : It is the one having the highest frequency...

Well i didnt draw the graph but i fink this information is enough. Just look the time which have the highest frequency...

(c) U only have to add the frequencies of the times above eight, that is, 7 + 4 + 2 = 13

Oh alright i kept on getting 4 because i thought mode and modal are the same thing, they are completely different right i mean mode is when a number keeps occuring, and modal is the highest frequency, am i right?

 

[zainabkhann]

Ok I solved it then...

thankyou, now im going to do the second part of the question and il let you know if if get the right answer, for lower bound , all i have to do is subtract the perimtere by 0.5 right?

 

[Minato112]

Yea somewhat...