All categories

2 years ago

18,18,15,18,18,24,21,21,24,14 which one is mean

Mathematics

1 answer:

julsineya [31]2 years ago

7 0

Answer:19.1

Step-by-step explanation: Put the numbers in order and then add the numbers and then divide it by how many number there is

14+15+18+18+18+18+21+21+24+24=191

191/10=19.1

You might be interested in

While watching a movie, Ned, Fred, and Jed shared a bowl of popcorn. Ned ate 1/2 of the popcorn, Fred ate 1/4 of the popcorn, an

Charra [1.4K]

1/8 bowl.
1/2 = 4/8
1/4 = 2/8
1/8 = 1/8

4/8+2/8+1/8 = 7/8
1 - 7/8 = 1/8

6 0

2 years ago

Read 2 more answers

Krutika is 11 years older than Gordon. Pip is 12 years younger than Gordon. If the total of their ages is 77, how old is the you

Vesnalui [34]

77 I hope it’s correct

7 0

2 years ago

W/3 + 6= 8<br><br>the w/3 is a fraction (and solve for w)

Mademuasel [1]

W is equal to 6 because 6/3=2 so 2+6=8

4 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

2 years ago

Please help me ASAP

Natasha2012 [34]

Is is b I am sure that is the right

7 0

2 years ago