miyoko23q3

2021-11-18

A.4.2569

B.4.4232

C.4.1509

D.4.0534

Mike Henson

Beginner2021-11-19Added 11 answers

Given data,

The table of the data is,

$$\begin{array}{|cc|}\hline x& f\left(x\right)\\ -0.1& 1.45\\ 0.2& 2.20\\ 0.4& 3.76\\ 0.5& 4.32\\ \hline\end{array}$$

Step 1

The Q notation table is,

$$\begin{array}{|cccc|}\hline {x}_{0}& {p}_{0}={Q}_{0.0}& & \\ {x}_{1}& {p}_{1}={Q}_{1.0}& {p}_{1.1}={Q}_{1.1}\\ {x}_{2}& {p}_{2}{Q}_{2.0}& {p}_{1.2}={Q}_{2.1}& {p}_{0.1.2}={Q}_{2.2}\\ {x}_{3}& 3.76& {p}_{2.3}={Q}_{3.1}& {p}_{1.2.3}{Q}_{3.2}\\ {x}_{4}& 4.32& {p}_{3.4}={Q}_{4.1}& {p}_{2.3.4}\\ \hline\end{array}$$

Step 2

Calculating the first degree approximation,

$Q}_{3.1}=\frac{(x-{x}_{2}){Q}_{3.0}-(x-{x}_{3}){Q}_{2.0}}{{x}_{3}-{x}_{2}$

$=\frac{(0-0.4)\cdot 4.32-(0-0.5)\cdot 3.76}{0.5-0.4}$

$=\frac{-0.4\cdot 4.32+0.5\cdot 3.76}{0.1}$

$=1.52$

$Q}_{2.2}=\frac{(x-{x}_{1}){Q}_{2.1}-(x-{x}_{2}){Q}_{1.1}}{{x}_{2}-{x}_{1}$

$=\frac{(0-0.2)\cdot 0.64-(0-0.4)\cdot 1.7}{0.4-0.2}$

$=\frac{-0.2\cdot 0.64+0.4\cdot 1.7}{0.2}$

$=2.76$

Step 3

The value of f (0) is 0.1432.

The sum of${Q}_{3.1}+{Q}_{2.2}+P\left(O\right)$ is,

${Q}_{3.1}{Q}_{2.2}+P\left(O\right)=1.52+2.76+0.1432$

$=4.4232$

Hence option Bis correct.

The table of the data is,

Step 1

The Q notation table is,

Step 2

Calculating the first degree approximation,

Step 3

The value of f (0) is 0.1432.

The sum of

Hence option Bis correct.