Question

Need help with linear approximation


Use linear approximation to estimate the following quantities. Choose a value of a to produce a small error.
1/203

Answer 1

I don't understand this question.

Answer 2

alright, im here boy

Answer 3

lemme firgure dis out, hold on.

Answer 4

ok thanks

Answer 5

You know Calculus?

Answer 6

10-16 (10 min.) Estimating a cost function.

1. Slope coefficient =

= 

= = $0.35 per machine-hour

Constant = Total cost – (Slope coefficient × Quantity of cost driver)

= $5,400 – ($0.35 × 10,000) = $1,900

= $4,000 – ($0.35 × 6,000) = $1,900

The cost function based on the two observations is
Maintenance costs = $1,900 + $0.35 × Machine-hours

2. The cost function in requirement 1 is an estimate of how costs behave within the relevant range, not at cost levels outside the relevant range. If there are no months with zero machine-hours represented in the maintenance account, data in that account cannot be used to estimate the fixed costs at the zero machine-hours level. Rather, the constant component of the cost function provides the best available starting point for a straight line that approximates how a cost behaves within the relevant range.
i hope this helps!!

Answer 7

what the hell is this? lol

Answer 8

or were you looking for more of a straight up answer?? hahah(:

Answer 9

Wat?? Tht makes no sense!

Answer 10

yes it does

Answer 11

I was kinda looking for an answer related to my question lol

Answer 12

thts straight off the internet lol :P

Answer 13

lol that was an example...

Answer 14

and no i did that myself

Answer 15

I no, but it's a confusing one. Maybe if u made it a bit more legible than he would understand lol

Answer 16

well lemme try to make this easier... ahahah! hold on!

Answer 17

can you help me find the linear approximation of 1/203

Answer 18

lol :)

Answer 19

i get the idea of linear approximation but how do I find an approximation of a constant?

Answer 20

A constant-factor approximation?

Answer 21

Approximate 1/.203.
y = -25(0.203) + 10 = -5.075 + 10 = -4.925

Answer 22

is that correct i think...

Answer 23

idk where did you get all these other numbers?

Answer 24

lol... you gotta do the work..

Answer 25

i give up :(

Answer 26

sorry, wished i couldve helped!

Answer 27

ummmmmmmmm okay thanks anyway

Answer 28

i got u one sec

Answer 29

ok thnks

Answer 30

You wish to find a "good" linear approximation of \(f(x)=\frac1x\) near \(x=203\) in order to approximate the value at \(x=203\) using relatively simple computation. How about we decide to pick our approximation at \(a=200\)?
\[y-f(a)=f'(a)[x-a]\\y-\frac1{200}=-\frac1{200^2}[x-200]\]
Thus we may approximate \(f(203)\approx y(203)=-\frac3{200^2}+\frac1{200}=\frac{200-3}{200^2}=\dots\)

Answer 31

ok I see what they're doing. Thanks, now it makes senses.