Mrs. Kemsley’s family always travels to a relative's

The softball team stopped for ice cream after a big game. five 2-scoop cones cost the same as three sundaes. If they bought nine

Savatey [412]

<span>Let the cost of 2-scoop cones be x, and the cost of sundaes be y.
The first statement tells us that 5x = 3y
The second statement tells us that 9x + 6y = 19.95
Then we need to calculate the value of y (the cost of a sundae).
From our first equation, y = 5x/3. Substituting this into the second equation:
9x + 6(5x/3) = 19.95
9x + 10x = 19.95
19x = 19.95
x = 1.05
Substituting back into the first equation: 5(1.05) = 3y
5.25 = 3y
y = 1.75
Therefore the cost of a sundae is $1.75.
</span>

3 0

3 years ago

Javier filled his gas tank with 8 gallons of unleaded gas for a cost of $15.12. How much will it cost to fill a 14-gallon tank?

tatuchka [14]

$26.46
I divided 15.12 by 8 to get the cost of just one gallon , which was $1.89. 14 x 1.89 = 26.46, which would ideally be the cost to fill a 14-gallon tank.

7 0

2 years ago

How many eights make a half

evablogger [386]

Answer:

Four, because 4/8 = 1/2 There are obviously 8 1/8ths in a whole so half of that amount is obviously 4 1/8ths.

Step-by-step explanation:

Gave above.

7 0

3 years ago

What’s the probability of getting 6 or more girls from 8 births?

Lemur [1.5K]

Answer:

3/4 or 6/8

Step-by-step explanation:

6/8 when simplified is 3/4. That is the probability over the total.

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

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

so we end up with

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

3 0

2 years ago