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2 years ago

15

What type of data distribution is shown on the graph? A bar graph has x on the x-axis, from 0 to 4, and P x (x) on the y-axis, f

rom 0 to 0.5 in 0.1 increments. The probability of 0 is 0.01, the probability of 1 is 0.25, the probability of 2 is 0.3, the probability of 3, is 0.4, and the probability of 4 is 0.02.

1 answer:

Leokris [45]2 years ago

8 0

Answer:

1.A

2.B

3.A

Step-by-step explanation:

edge 2021 yw :)

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Answer plz I need it !! Plz

adelina 88 [10]

4 is b p=2.5n
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2 years ago

Bobby made cookies. the circumference of cookies are 6 cm. what is the diameter of the cookies?

Harlamova29_29 [7]

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2 years ago

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

2 years ago

PLEASE HELP! Thank you so much! I’m

Kay [80]

Convert both fractions into a larger fraction that applies to both. Here that would be 2 $\frac{2}{10}$ and 2 $\frac{1}{2}$ , which is the same as 2 $\frac{5}{10}$

That means M is 2 $\frac{3}{10}$

hope that helps!

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2 years ago

Find the payment of $435 with a 6% sales tax.

Kobotan [32]

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1 year ago