Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
The data below shows the average number of text messages sent daily by a group of people: 7, 8, 4, 7, 5, 2, 5, 4, 5, 7, 4, 8, 2,
enot [183]
It all depends. You've given us an incredibly vague question.
The outlier could be a number that's low or quite high. Also, outliers
shouldn't really contribute towards the value of the mean, median or
range related to a group of data.
They are called outliers because they are bizarre results or numbers
and should be detached from groups of data. Outliers by definition
are abnormalities or anomalies.
I'd say outliers don't really change anything, unless you actually want
to give them credibility or weight.
Large outliers can inflate the value of means, medians and ranges.
Small outliers will invariably deflate the value of means and medians.
9/11 is rounded to 1. It rounds up because 9 is higher than 5, which is the middle of 11.
Answer:
4.3
Step-by-step explanation:
look at the picture .......
Answer:
7.5
Step-by-step explanation:
cause you can divide the 60 with the 8
