Giving away points, because I'm leaving :3

C) A researcher has extracted two samples from the same target population, sample A and sample B. Sample A has a mean of 25 kg w

aleksandrvk [35]

Answer:

dsgberhyujxgkloyjnhbgfvbhnjklkjhbgfdfghjkiopolkjhgfew

Step-by-step explanation:

3 0

3 years ago

Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI

cricket20 [7]

Answer:

The system has infinitely many solutions

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

Switch any two rows
Multiply a row by a nonzero constant
Add one row to another

To solve the following system

$\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}$

Step 1: Transform the augmented matrix to the reduced row echelon form

$\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]$

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

which corresponds to the system

$\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}$

The system has infinitely many solutions.

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

7 0

3 years ago

Which quadrilateral below could have four different side lengths?

makvit [3.9K]

Answer:

c. trapezoid

Step-by-step explanation:

4 0

3 years ago

1. Find the length a.

stellarik [79]

Use the Pythagorean Theorem to find the missing lengths of the sides of a right triangle. triangle that has an opposite side of length 3 and a hypotenuse of length 4. Determining all of the side lengthsand angle measures of a right triangle is Let's look at how to do this when you're given one side lengthand one acute

7 0

3 years ago

Determine whether the following statements are true and give an explanation or a counter example. a. The Trapezoid Rule is exact

Norma-Jean [14]

Answer:

statement is TRUE

statement is FALSE

statement is TRUE

Step-by-step explanation:

(a)

By using the Trapezoidal Rule, the definite integral can be computed by applying linear interpolating formula on each sub interval, and then sum-up them, to get the value of the integral

So, in computing a definite integral of a linear function, the approximated value occurred by using Trapezoidal Rule is same as the area of the region. Thus, the value of the definite integral of a linear function is exact, by using the Trapezoidal Rule.

Therefore, the statement is TRUE

(b) Recollect that for each rule of both the midpoint and trapezoidal rules, the number of sub-internals, n increases by a factor of a. then the error decreases by a factor of a^2.

So, for the midpoint rule, the number of sub-intervals, n is increased by a factor of 3, then the error is decreased by a factor of 32 = 9, not 8. Therefore, the statement is FALSE

(c) Recollect that for each rule of both the midpoint and trapezoidal rules, the number of sub-internals, n increases by a factor of a. then the error decreases by a factor of a^2.

So, for the trapezoidal rule, the number of sub-internals, n is increased by a factor of 4. then the error is decreased by a factor of 42 = 16

Therefore, the statement is TRUE

3 0

3 years ago