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Anvisha [2.4K]
2 years ago
8

A jar contains four blue marbles, three red marbles, and two white marbles. What is the probability that a marble randomly selec

ted from the jar will be a white marble?
A. 11.1%
B. 22.2%
C. 28.6%
D. 50.0%
Mathematics
2 answers:
Darya [45]2 years ago
3 0

Answer:

22.2%

Step-by-step explanation:

It is 22.2% of 100 %.We first know that there are 2 white marbles.Ad a zero and that's what you get.

Masja [62]2 years ago
3 0

Answer:

B.

Step-by-step explanation:

100÷9=11.1

11.1×2=22.2

Mark me brainliest

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3/4 + (-1/2)

= 3/4 - 1/2

= (3 - 2)/4

= 1/4

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2 years ago
Jennifer has 7 marbles in her bag. She has 4 blue, 1 white, 1 pink, and 1 yellow. She chooses one marble without looking. What i
UNO [17]

Answer:

all of the above

Step-by-step explanation:

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scoundrel [369]
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Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

3 0
2 years ago
Maya deposits $5000 into a checking account that pays 0.75% annual interest compounded monthly. What will be the balance after 8
Nana76 [90]

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 0.75\%\to \frac{0.75}{100}\dotfill &0.0075\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &8 \end{cases} \\\\\\ A=5000\left(1+\frac{0.0075}{12}\right)^{12\cdot 8}\implies A=5000(1.000625)^{96}\implies A\approx 5309.08

5 0
1 year ago
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