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3 years ago

Find the median of the following number 40,30,32,39,34,35,35,37

Mathematics

2 answers:

vova2212 [387]3 years ago

8 0

Final Answer:

35

Step-by-step explanation:

-median is the middle number in the data set-

arrange the data set from least to greatest:

40, 30, 32, 39, 34, 35, 35, 37 ⇒ 30, 32, 34, 35, 35, 37, 39, 40

we start from the lowest number and the highest number and work your way to the middle:

30, 32, 34, 35, 35, 37, 39, 40

notice that there is no middle number:

30, 32, 34, 35, ?, 35, 37, 39, 40

so what we do is add both 35's and then divide it by 2

35 + 35 = 70 ÷ 2 = 35

median: 35

aleksandr82 [10.1K]3 years ago

3 0

Terms given

40,35,32,39,34,35,35,37

No of terms=8

We know

$\boxed{\sf Mean=\dfrac{Sum\;of\:terms}{No\;of\:terms}}$

$\\ \sf\longmapsto Mean=\dfrac{40+35+32+39+34+35+35+37}{8}$

$\\ \sf\longmapsto Mean=\dfrac{270}{8}$

$\\ \sf\longmapsto Mean=35$

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Answer:

Let $t_i$ be the time for the $i$ th task.

We know these times have a certain structure:

Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

$t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6$

The second task takes 1 second $t_2=1$

The fourth task takes 10 seconds $t_4=10$

So, we have the following system of equations:

$t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10$

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

$\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

Subtract row 2 from row 1 $\left(R_1=R_1-R_2\right)$ .
Subtract row 2 from row 3 $\left(R_3=R_3-R_2\right)$ .
Add row 3 to row 2 $\left(R_2=R_2+R_3\right)$ .
Multiply row 3 by −1 $\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)$ .
Add row 4 multiplied by $\frac{3}{2}$ to row 1 $\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)$ .
Subtract row 4 from row 3 $\left(R_3=R_3-R_4\right)$ .

Here is the reduced echelon form for the augmented matrix.

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

c) The additional rows are

$\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right$

and the augmented matrix is

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]$

d) To solve the system you must use these row operations.

Subtract row 1 from row 6 $\left(R_6=R_6-R_1\right)$ .
Subtract row 2 from row 6 $\left(R_6=R_6-R_2\right)$ .
Subtract row 3 from row 6 $\left(R_6=R_6-R_3\right)$ .
Swap rows 5 and 6.
Add row 5 to row 3 $\left(R_3=R_3+R_5\right)$ .
Multiply row 5 by 2 $\left(R_5=\left(2\right)R_5\right)$ .
Subtract row 6 multiplied by 1/2 from row 1 $\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)$ .
Add row 6 multiplied by 1/2 to row 3 $\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)$ .

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]$