Binomial Theory: "A square golden slab was measured 2% too large...."

[JimCrown]

I am completely stumped at how to approach and answer this question.

Use binomial theory to calculate error. A square golden slab was measured 2% too large. This measurement was used to calculate the amount of slabs needed for a square patio. The patio is required, by the customer, to be exactly 6m^2 (6 metres squared) and the slabs are 0.36m^2 (0.36 metres squared).

What would the resulting error in the area be?
What are your conclusions with respect to laying the patio?

Would I just put the 0.36m^2 in the theorem/formula?

 

[Steven G]

I am completely stumped at how to approach and answer this question.

Use binomial theory to calculate error. A square golden slab was measured 2% too large. This measurement was used to calculate the amount of slabs needed for a square patio. The patio is required, by the customer, to be exactly 6m^2 (6 metres squared) and the slabs are 0.36m^2 (0.36 metres squared).

What would the resulting error in the area be?
What are your conclusions with respect to laying the patio?

Would I just put the 0.36m^2 in the theorem/formula?

Would I just put the 0.36m^2 in the theorem/formula? Maybe? I might be able to answer your question if you stated what theorem or formula you were talking about.

What would the resulting error in the area be? What is the area without any error (you really should know this). By how much was the area off by? ....

 

[JimCrown]

I think it is just binomial expansion but, I do not have the area. I am expected to work this out (I am guessing it is 6m^2). Do I just find the 2% of 0.36m^2 which is 0.002592 and take that away with the 2% from one

(1 - 0.02)(1 - 0.002592)
(1 - 0.022592)
= 0.977408
= 97.74 Percentage error?

Or do I just divide 6^2 by 0.36^2 and find the area that way. I AM completely lost.

 

[JimCrown]

Would I just put the 0.36m^2 in the theorem/formula? Maybe? I might be able to answer your question if you stated what theorem or formula you were talking about.

What would the resulting error in the area be? What is the area without any error (you really should know this). By how much was the area off by? ....

I have posted a response above

 

[Ishuda]

I am completely stumped at how to approach and answer this question.

Use binomial theory to calculate error. A square golden slab was measured 2% too large. This measurement was used to calculate the amount of slabs needed for a square patio. The patio is required, by the customer, to be exactly 6m^2 (6 metres squared) and the slabs are 0.36m^2 (0.36 metres squared).

What would the resulting error in the area be?
What are your conclusions with respect to laying the patio?

Would I just put the 0.36m^2 in the theorem/formula?

Let us work through an example using a 6 meter length rather than a 6 square meter area.


First, I would ask measured 2% too large in length or measured 2% too large in area? Assuming the measurement of the length was 2% too large and we want the error in the area, we have the wanted area
A₀= 6² = 36
and the binomial theorem gives the actual area
A₁ = [6*1.02]² = [6 + 0.12]² ~ 6² 0.12⁰ + 2 6¹ 0.12¹ + 6⁰ 0.12² = 37.4544 m²
The error is
A1 - A0 = 1.4544 m²
or the area was about 4% too large.

Suppose it had been measured 2% to large in area. That would mean that
e = 6 - \(\displaystyle \sqrt{36*1.02}\, =\, 6\, [1\, -\, \sqrt{1\, +\, .02}\)]
where e is the error in length. We could then use use the binomial theorem to determine a value [an approximate value] of e.