All categories

3 years ago

Why are the mode median and mean call measures of the center?

Mathematics

2 answers:

Yuliya22 [10]3 years ago

7 0

Answer:

Step-by-step explanation:

These are statistics that describe the center of a distribution: mean, median, mode. They measure central _____.

The answer is Tendency.

Hope this helps!!!!!!!!

Contact [7]3 years ago

6 0

Because they are in the center.

You might be interested in

Meg loves to read. Her favorite author is Dr. Seuss. For every 11 books Meg owns 8 are Dr. Seuss books. If Meg has 40 Dr. Seuss

ANTONII [103]

Answer:

Step-by-step explanation:

if the ratio is 8:11 then 40:x

Cross multiply

8 40

11 x

8x=440

And divide by 8

x=55. She has 55 books

8 0

3 years ago

After two plays the wildcats gained a total of 16 yards. If X represents the number of yards for play one, and Y represents the

Taya2010 [7]

The equation for this problem would be:
x+y = 16
Rearranging the equation, it can be expressed as:
y = 16 - x

To obtain ordered pairs, replace random values of x up to 16 to find the value of y. The ordered pairs are tabulated and the graph is shown in the attached picture.

8 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

A collection of quarters, dimes, and nickels is worth $13.30. If the ratio of quarters to

RideAnS [48]

Answer:

105

Step-by-step explanation:

35 quarters ($8.75)

21 dimes ($2.10)

49 nickels ($2.45)

7 0

2 years ago

( 7x + 1/2) + ( 7x - 7 and 1/2) Simplify

pishuonlain [190]

Answer:

The answer will be 7(2x - 1)

Step-by-step explanation:

1) Convert 7 and 1/2 to improper fraction. Use this rule: a b/c = ac + b/c

$7x + \frac{1}{2} + 7x - \frac{7 \times 2 + 1}{2}$

2) Simplify 7 × 2 to 14 .

$7x + \frac{1}{2} + 7x - \frac{14 + 1}{2}$

3) Simplify 14 + 1 to 15.

$7x + \frac{1}{2} + 7x - \frac{15}{2}$

4) Collect like terms.

$(7x + 7x) + ( \frac{1}{2} - \frac{15}{2} )$

5) simplify.

$14x - 7$

6) Factor out the common term 7.

$7(2x - 1)$

Therefor, the answer is 7( 2x - 1 ).

7 0

2 years ago