Ask question

Ask question

All categories

2 years ago

7

If Jayna's shuffler is working properly and randomizing the playlist after each song, what is the probability of playing a slow

song?
ANSWER IS 33.33%

1 answer:

marta [7]2 years ago

4 0

Answer:

1 song is a slow song, 2 songs are not

Step-by-step explanation:

that would give you 1/3 probability = 33.3333333333333%

You might be interested in

PLEASE HELP ME! I NEED IT!

lesya [120]

The second choice.................................

7 0

2 years ago

Point Q is located at (-4, 6). Point R is located at (8, 6).

Bess [88]

Step-by-step explanation:

Hi there!

Given;

Point Q is located at (-4, 6). Point R is located at (8, 6).

Note: Use distance formula.

Now;

$(d) = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

Keep all values;

$(d) = \sqrt{ {(8 + 4)}^{2} + ( {6 - 6)}^{2} }$

Simplify;

$(d) = \sqrt{( {12)}^{2} }$

Therefore, the distance is 12 units.

<u>Hope</u><u> it</u><u> helps</u><u>!</u>

8 0

2 years ago

What is the answer to x? Help ASAP

Brut [27]

Answer:

12

Step-by-step explanation:

2x + 24° = 4x ( being vertically opposite angles)

4x - 2x = 24°

2x = 24°

x = 24° / 2

x = 12°

Hope it will help :)

3 0

2 years ago

5 less than half a number, x<br><br> please help i will give you the brainliest answer

Karo-lina-s [1.5K]

Answer: 1/2x -5

Step-by-step explanation:

Half of x , less 5. (5 less just means subtract)

5 0

2 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

If

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

2 years ago