Step-by-step explanation:
The Taylor series expansion is:
Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!
f(x) = 1/x, a = 4, and n = 3.
First, find the derivatives.
f⁽⁰⁾(4) = 1/4
f⁽¹⁾(4) = -1/(4)² = -1/16
f⁽²⁾(4) = 2/(4)³ = 1/32
f⁽³⁾(4) = -6/(4)⁴ = -3/128
Therefore:
T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!
T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³
f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0. So we can eliminate the top left option. That leaves the other three options, where f(x) is the blue line.
Now we have to determine which green line is T₃(x). The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).
The bottom right graph is the only correct option.
Answer:
the x² test statistic 13.71
Option a) 13.71 is the correct answer.
Step-by-step explanation:
Given the data in the question;
Feeder 1 2 3 4
Observed visits; 
60 90 92 58
data sample = 300
Expected 
= 300 / 4 = 75
the x² test statistic = ?
= ∑[ ( - )²/]
= [ (60 - 75)² / 75 ] + [ (90 - 75)² / 75 ] + [ (92 - 75)² / 75 ] + [ (58 - 75)² / 75 ]
= [ 3 ] + [ 3 ] + [ 3.8533 ] + [ 3.8533 ]
= 13.7066 ≈ 13.71
Therefore, the x² test statistic 13.71
Option a) 13.71 is the correct answer.
I believe the answer is “A”
5.4 yards for each dress. This number lies between 5 and 6
