All categories

3 years ago

in chemistry class Andrew has earned scores of 64 69 and 73 on three tests he must maintain an average of 72

1 answer:

6 0

You haven't told me what the question is. But I put the mouse
to my forehead, closed my eyes, took a deep breath, and I could
see it shimmering in my mind's eye. It was quite fuzzy, but I think
the question is

   "What score does Andrew need on the next test
   in order to raise his average to 72% ?"

The whole experience drew an incredible amount of energy
out of me, and the mouse is a total wreck. So we'll just go ahead
and answer that one. I hope it's the correct question.

The average score on 4 tests is

   (1/4) (the sum of all the scores) .

In order for Andrew to have a 72% average on 4 tests,
the sum of the 4 scores must be

   (4) x (72%) = 288% .

Out of that total that he needs, he already has

(64% + 69% + 73%) = 206%

on the first three tests.

So in order to average 72% for all 4 tests,
he'll need to score

   (288% - 206%) = 82%

on the fourth one.

You might be interested in

What is the solution of the following system of equation: 2X+Y=17 AND Y=X+14 ?

Inessa [10]

Answer:

I do't know how to solve this.

Step-by-step explanation:

6 0

3 years ago

It was reported that 23% of U.S. adult cellphone owners called a friend for advice about a purchase while in a store. If a sampl

leva [86]

Answer:

$P(X=7)$

And using the probability mass function we got:

$P(X=7)=(15C7)(0.23)^7 (1-0.23)^{15-7}=0.0271$

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=0.23)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find the following probability:

$P(X=7)$

And using the probability mass function we got:

$P(X=7)=(15C7)(0.23)^7 (1-0.23)^{15-7}=0.0271$

5 0

3 years ago

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 $n$ th 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 $U\sim N(n,1)$ :