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-BARSIC- [3]
3 years ago
9

What's the Constant of Proportianlity/Unit Rate?​

Mathematics
1 answer:
Leto [7]3 years ago
3 0

Answer:

1/3 cup of juice per cup of berries

Step-by-step explanation:

The table shows that if we go from 0 to 4 cups (a "run" of 4 cups), the juice output is 1 1/3 (or 4/3) (a "rise" of 4/3 cups), and so the desired constant of proportionality is the unit rate

(4/3) cups

------------------------ = 1/3 cup of juice per cup of berries

  4 cups berries

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The table represents the height of a rock that is dropped, h(t), in meters, after t seconds. (See picture.)
Misha Larkins [42]

Answer: Between <u>2</u> and <u>2.5</u> seconds

How am I getting those values? Those values are selected from the 't' column on the left side of the table. These are time values in seconds. Note how t = 2 corresponds to h = 0.4; while t = 2.5 corresponds to h = -10.6

The change in sign for the height, from positive to negative, means that the height h must be zero at some point between t = 2 seconds and t = 2.5 seconds. We don't know where exactly, but we know that h = 0 at least once in this interval. This is because the height is a continuous function. There are no jumps or gaps in the height (the object can't teleport or anything)

3 0
3 years ago
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60 ft/s = ____km/h I don't know how to convert feet to kilometers.
kondor19780726 [428]
Well to convert feet into km you have to know that 1 feet = 0.0003048 km

knowing that you multiply the km by 60 which gives you..... 0.018288 km 

7 0
3 years ago
Trignometry help what's the answer?
butalik [34]
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 See attached picture for solution:

7 0
3 years ago
Which expression is equivalent to 2^4 ⋅ 2^−7?
yaroslaw [1]
Remember
(x^m)(x^n)=x^{m+n}
and
x^{-m}=\frac{1}{x^m}

so
(2^4)(2^{-7})=
2^{4-7}=
2^{-3}=
\frac{1}{2^3}=
\frac{1}{8}
8 0
3 years ago
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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
3 years ago
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