All categories

3 years ago

What is the range of the following questions .406 .407 .408 .405 .411 .406

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

5 0

Answer: 407 is the range

Step-by-step explanation: 406 + 407 + 408 + 405 + 411 + 406 = 2,443

and count how many numbers there are. 6 then you divide 2,443 with 6 = 407

WOOOAH

You might be interested in

A construction company requires 25,400 bricks

maks197457 [2]

25,908 bricks must be ordered

8 0

3 years ago

A circle has a circumference of 25.12 inches what is the diameter of the circle

BigorU [14]

Answer:

Step-by-step explanation:

Recall these two equations for circumference and diameter.

Circumference = 2*pi*radius

Diameter = radius*2

25.12 = 2*pi*radius

Radius = 4

Diameter = 4*2 => 8

Thus the diameter is 8

4 0

3 years ago

What is the answer please??

xenn [34]

Answer:

Answer is n/2

Step-by-step explanation:

I started with the constant n

In second step I divided the n by 2

$answer \: is \: \: \frac{n}{2}$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

5 0

3 years ago

Read 2 more answers

What is the oblique asymptote of the function f(x) x^2+x-2/x+1

Mamont248 [21]

$f(x)=\displaystyle\frac{x^2+x-2}{x+1}=\frac{x(x+1)-2}{x+1}=x-\frac{2}{x+1}
$

$\displaystyle\lim_{x\to\infty}\left\{f(x)-x \right\}=\lim_{x\to\infty}\frac{2}{x+1}=\lim_{x\to\infty}\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}
=0$

Consequently, the limit of $f(x)$ as x approaches infinity is $x$ .

In other words, $f(x)$ approaches the line y=x,

so oblique asymptote is y=x.

I'm Japanese, if you find some mistakes in my English, please let me know.

3 0

3 years ago

A researcher reports survey results by stating that the standard error of the mean is 20. The population standard deviation is 5

Mekhanik [1.2K]

multiply 20 by 500. it should help you with the problem.

5 0

3 years ago