2 questions (1 Viewer)

[YourLocalDumbAss]

1. Is Simpson’s rule still applicable with the current HSC Syllabus?
2. Is it the same as the trapezoidal rule or different, if so what’s the formula for it?

Source 2018 Standard Maths HSC Exam.

 

[specificagent1]

Trapezoidal rule is different. Simpson's rule is no longer part of the adv course i cant speak for the standard

 

[YourLocalDumbAss]

Trapezoidal rule is different. Simpson's rule is no longer part of the adv course i cant speak for the standard

Thank you for telling me it’s different.
Could you tell me the formula and what means what

 

[jimmysmith560]

Thank you for telling me it’s different.
Could you tell me the formula and what means what

There is no mention of Simpson's rule in the Mathematics Standard syllabus document (as opposed to the Trapezoidal rule), which possibly indicates that Simpson's rule is no longer assessed in Mathematics Standard.

To answer your question, the approximation using Simpson's rule is as follows:



I hope this helps!

 

[YourLocalDumbAss]

There is no mention of Simpson's rule in the Mathematics Standard syllabus document (as opposed to the Trapezoidal rule), which possibly indicates that Simpson's rule is no longer assessed in Mathematics Standard.

To answer your question, the approximation using Simpson's rule is as follows:

View attachment 32845

I hope this helps!

Thanks Jimmy

 

[CM_Tutor]

Simpson's Rule is definitely no longer in the syllabus.

It often gives more accurate approximations than the Trapezoidal Rule as it approximates using parabola through groups of three points (ie. using two strips).

 

[YourLocalDumbAss]

Simpson's Rule is definitely no longer in the syllabus.

It often gives more accurate approximations than the Trapezoidal Rule as it approximates using parabola through groups of three points (ie. using two strips).

Thank you

 

[YourLocalDumbAss]

How about this question, is this also not in the syllabus?

If not, how do you calculate it?

 

[Lith_30]

If not, how do you calculate it?

1 byte = 8 bits (1B=8b), therefore 495MB=3960Mb
Time taken to download would be seconds

Not sure if it is in the syllabus though.

Edit:
Thanks for CM_Tutor and jimmysmith560 for pointing out my error.

 

[jimmysmith560]

View attachment 32849
How about this question, is this also not in the syllabus?

If not, how do you calculate it?

There is a section within the Mathematics General syllabus (old syllabus) that covers specific terms relevant to this question:



There is no mention of this in the Mathematics Standard syllabus (current syllabus), which also indicates that this is likely to no longer be part of the syllabus.

The working out for this answer is as follows:

 

[harrowed2]

View attachment 32849
How about this question, is this also not in the syllabus?

If not, how do you calculate it?

No longer in the Syllabus

 

[CM_Tutor]

1 byte = 8 bits (1B=8b), therefore 495MB=3960Mb
Time taken to download would be seconds

Not sure if it is in the syllabus though.

This calculation is incorrect as even though 1 byte = 8 bits, 1 megabyte is not 8 megabits

1 megabyte = 1024 kilobytes = 2¹⁰ kilobytes

1 kilobyte = 1024 bytes = 2¹⁰ bytes

Thus, 1 megabyte = 2²⁰ bytes = 2²⁰ x 8 bits = 2²³ bits

​

which matches the answer that @jimmysmith560 posted earlier

 

[Lith_30]

This calculation is incorrect as even though 1 byte = 8 bits, 1 megabyte is not 8 megabits

1 megabyte = 1024 kilobytes = 210 kilobytes

1 kilobyte = 1024 bytes = 210 bytes

Thus, 1 megabyte = 220 bytes = 220 x 8 bits = 223 bits

​

which matches the answer that @jimmysmith560 posted earlier

Edited my above post.