All categories

2 years ago

Find the mean of 3,2,4,6,5,6,1,7,9,8,10

Mathematics

1 answer:

LenaWriter [7]2 years ago

5 0

Answer:

$5\frac{6}{11}$

Step-by-step explanation:

Number of Values = 11.

Mean = Sum of values / Number of Values

= $\frac{3+2+4+6+5+6+1+7+9+8+10}{11} \\=\frac{61}{11} \\= 5\frac{6}{11}$

You might be interested in

How to write 11001195 in words

avanturin [10]

Answer:

Eleven million one thousand one hundred ninety-five

6 0

3 years ago

Read 2 more answers

Can someone answer this

bezimeni [28]

Answer:

Step-by-step explanation:

x 6 a

8 48 c

-4 b 20

Let the unknown numbers of the multiplication grid are a, b and c.

1). 6 × 8 = 48

2). (-4)×6 = b

b = -24

3). (-4) × a = 20

a = -5

4). 8 × a = c

8 × (-5) = c

c = -40

Therefore, missing in the given multiplication grid are,

x 6 -5

8 48 -40

-4 -24 20

6 0

3 years ago

1. -7/9 • 3/5

ASHA 777 [7]

The answer to the question

5 0

3 years ago

Find the ratio of x to y x/5 = 2/3 = 5/y

BigorU [14]

X=2
-- ---
5 3
Cross multiply:
3x=10
x=10/3

2=5
-- ---
3 y
2y=15
y=15/2
10/3=15/2 :)

3 0

3 years ago

Read 2 more answers

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago

Find the mean of 3,2,4,6,5,6,1,7,9,8,10​

Find the mean of 3,2,4,6,5,6,1,7,9,8,10