What is the mode of this data set? {7, 10, 12, 9, 12, 4, 13, 14}

Mathematics

2 answers:

Shkiper50 [21]2 years ago

8 0

5; 7; 19; 24; 10; 17; 21; 6; 22; 5; 9

Arrange: 5; 5; 6; 7; 9; 10; 17; 19; 21; 22; 24

( 5+5+6+7+9+10+17+19+21+22+24 ) / 11

Mean = 145 / 11

Mean = 13,18

Mode = 5

(5 is the only value that appears most often)

Median = 10

(10 is the middle value)

Range = 24 - 5

=19

( Range is the difference between largest and smallest values)

trapecia [35]2 years ago

3 0

The mode of this set of data is 12

You might be interested in

Y =2/3x + 20 when x = 21

nydimaria [60]

Answer:

Step-by-step explanation:

First we need to do 2/3*21, which equals 14 as 3 and 21 can be simplified to 1 and 7. 7 * 2/1= 14. Then we add the 20 to 14, 20+14= 34

6 0

2 years ago

Read 2 more answers

a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri

SSSSS [86.1K]

In the $x$ direction we consider the $m=2$ subintervals [0, 1] and [1, 2] (each with length 1), while in the $y$ direction we consider the $n=2$ subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of $R$ are (1, 0), (2, 0), (1, 2), (2, 2).

Let $f(x,y)=7x+5y^2$ . The volume of the solid is approximately

$\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}$

###

More generally, the lower-right-corner Riemann sum over $m=\mu$ and $n=\nu$ subintervals would be

$\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)$