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Sauron [17]
2 years ago
8

  The profit on the sale of a small bag of popcorn is $1, while the profit on the sale of a medium bag of popcorn is $2. The Gle

e Club expects to sell fewer than 100 total bags of popcorn, but they hope to make more than $100 in profit. The system that represents this is s + m<100 and s+2m>100. 
Mathematics
1 answer:
Brrunno [24]2 years ago
7 0

Let number of small bags = s

Let number of medium bags = m

Given profit on 1 small bag = $1

then profit on s bags = 1s

Given profit on 1 medium bag = $2

then profit on m bags = 2m

Now total profit from both type of bags is 1x+2y

profit should be more than $100 so we get inequality

1s+2m>100

Total number of bags are fewer than 100 so we get

s+m<100

which are same as inequalities you found.

Hence your answer is correct.

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Tom [10]

Answer:

Answer D is correct

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2 years ago
4.52 Random variables X and Y follow a joint distribution f(x, y) = 2, 0 &lt; x ≤ y &lt; 1, 0, otherwise. Determine the correlat
VMariaS [17]

First, you'll need to find the marginal distributions of X,Y. By the law of total probability,

P(X=x)=\displaystyle\sum_yP(X=x)P(Y=y)

which translates to

f_X(x)=\displaystyle\int_x^1f_{X,Y}(x,y)\,\mathrm dy=\begin{cases}2(1-x)&\text{for }0

Similarly,

f_Y(y)=\displaystyle\int_0^yf_{X,Y}(x,y)\,\mathrm dx=\begin{cases}2y&\text{for }0

Compute the expectations for both random variables:

E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\int_0^12y^2\,\mathrm dy=\frac23

Compute the variances and thus standard deviations:

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16

\implies V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}\implies\sqrt{V[X]}=\dfrac1{3\sqrt2}

E[Y^2]=\displaystyle\int_{\infty}^\infty y^2f_Y(y)\,\mathrm dy=\int_0^12y^3\,\mathrm dy=\frac12

\implies V[Y]=\dfrac12-\left(\dfrac23\right)^2=\dfrac1{18}\implies\sqrt{V[Y]}=\dfrac1{3\sqrt2}

Compute the covariance:

\operatorname{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]

We have

E[XY]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^1\int_0^y2xy\,\mathrm dx\,\mathrm dy=\frac14

and so

\operatorname{Cov}[X,Y]=\dfrac14-\dfrac13\dfrac23=\dfrac1{36}

Finally, the correlation:

\operatorname{Corr}[X,Y]=\dfrac{\operatorname{Cov}[X,Y]}{\sqrt{V[X]}\sqrt{V[Y]}}=\dfrac{\frac1{36}}{\left(\frac1{3\sqrt2}\right)^2}=\dfrac12

3 0
2 years ago
Explain how a personal budget relates to and takes into account your personal financial goals.
nexus9112 [7]

what? what subject is it

6 0
2 years ago
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Katie is watching a helicopter flying at an altitude of 5 miles. If the angle of elevation is pi/3 , the distance between Katie
Art [367]
From SOH CAH TOA, you know that
.. Sin = Opposite/Hypotenuse
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3 0
2 years ago
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svp [43]

Answer:

0.0055

Step-by-step explanation:

This is a binomial probabilty distribution problem.

The formula is:

P(x=r)=nCr* p^{r}q^{n-r}

Where

n is the total number of trials

r is the number of trials we want

p is the probabilty of success

q is 1 - p [probability of failure]

Here, we take 16 people, so n = 16

we want probability of hallf will have high blood pressure, that is 16*0.5 = 8

so r = 8

p is probabilty of that happening, which is 20% , p = 0.2

q = 1 - 0.2 = 0.8

Now putting into formula, we have:

P(x=8)=16C8* (0.2)^{8}(0.8)^{16-8}\\=0.0055

Hence

probability one half will have high blood pressure is 0.0055

5 0
2 years ago
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