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Neko [114]
2 years ago
8

A chemist is researching different sustainable fuel sources. She is currently working with benzene, which must be in liquid form

for her to successfully conduct her
research. The boiling point of benzene is 176° F, and the freezing point is 42° F

Part A: Write an inequality to represent the temperatures the benzene must stay between to ensure it remains liquid. (2 points)

Part B: Describe the graph of the inequality completely from Part A. Use terms such as open/closed circles and shading directions. Explain what the solutions to the
inequality represent. (4 points)

Part C: In February, the building's furnace broke and the temperature of the building fell to 20° F Would the chemist have been able to conduct her research with
benzene on this day? Why or why not? (4 points)
Mathematics
1 answer:
White raven [17]2 years ago
7 0

Answer:

Quieres ser mi novia.. Hermosa.. MI estrella fugaz

Step-by-step explanation:

❤️❤️❤️❤️❤️❤️:-):-):-):-)amorsito...,,,,,,, - - - - ////

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How to put 33 into 4:7
Tema [17]
4+7 = 11
so
33 / 11 = 3

4 x 3 = 12
7 x 3 = 21

so answer
4:7 = 12:21
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3 years ago
Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the
fenix001 [56]

Answer:

(a): Marginal pmf of x

P(0) = 0.72

P(1) = 0.28

(b): Marginal pmf of y

P(0) = 0.81

P(1) = 0.19

(c): Mean and Variance of x

E(x) = 0.28

Var(x) = 0.2016

(d): Mean and Variance of y

E(y) = 0.19

Var(y) = 0.1539

(e): The covariance and the coefficient of correlation

Cov(x,y) = 0.0468

r \approx 0.2657

Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

P(x) = \sum\limits^{}_y\ P(x,y)

So:

P(0) = P(0,0) + P(0,1)

P(0) = 0.63 + 0.09

P(0) = 0.72

P(1) = P(1,0) + P(1,1)

P(1) = 0.18 + 0.10

P(1) = 0.28

Solving (b): Marginal pmf of y

This is calculated as:

P(y) = \sum\limits^{}_x\ P(x,y)

So:

P(0) = P(0,0) + P(1,0)

P(0) = 0.63 + 0.18

P(0) = 0.81

P(1) = P(0,1) + P(1,1)

P(1) = 0.09 + 0.10

P(1) = 0.19

Solving (c): Mean and Variance of x

Mean is calculated as:

E(x) = \sum( x * P(x))

So, we have:

E(x) = 0 * P(0)  + 1 * P(1)

E(x) = 0 * 0.72  + 1 * 0.28

E(x) = 0   + 0.28

E(x) = 0.28

Variance is calculated as:

Var(x) = E(x^2) - (E(x))^2

Calculate E(x^2)

E(x^2) = \sum( x^2 * P(x))

E(x^2) = 0^2 * 0.72 + 1^2 * 0.28

E(x^2) = 0 + 0.28

E(x^2) = 0.28

So:

Var(x) = E(x^2) - (E(x))^2

Var(x) = 0.28 - 0.28^2

Var(x) = 0.28 - 0.0784

Var(x) = 0.2016

Solving (d): Mean and Variance of y

Mean is calculated as:

E(y) = \sum(y * P(y))

So, we have:

E(y) = 0 * P(0)  + 1 * P(1)

E(y) = 0 * 0.81  + 1 * 0.19

E(y) = 0+0.19

E(y) = 0.19

Variance is calculated as:

Var(y) = E(y^2) - (E(y))^2

Calculate E(y^2)

E(y^2) = \sum(y^2 * P(y))

E(y^2) = 0^2 * 0.81 + 1^2 * 0.19

E(y^2) = 0 + 0.19

E(y^2) = 0.19

So:

Var(y) = E(y^2) - (E(y))^2

Var(y) = 0.19 - 0.19^2

Var(y) = 0.19 - 0.0361

Var(y) = 0.1539

Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

COV(x,y) = E(xy) - E(x) * E(y)

Calculate E(xy)

E(xy) = \sum (xy * P(xy))

This gives:

E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)

E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1

E(xy) = 0+0+0 + 0.1

E(xy) = 0.1

So:

COV(x,y) = E(xy) - E(x) * E(y)

Cov(x,y) = 0.1 - 0.28 * 0.19

Cov(x,y) = 0.1 - 0.0532

Cov(x,y) = 0.0468

The coefficient of correlation is then calculated as:

r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}

r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}

r = \frac{0.0468}{\sqrt{0.03102624}}

r = \frac{0.0468}{0.17614266944}

r = 0.26569371378

r \approx 0.2657 --- approximated

8 0
2 years ago
In a one-month period, a convenience store
Vera_Pavlovna [14]
So, youd start by adding the two sales together to get total sales. Then you would divide that number by the gas sales to get the percent

65560+36740 = x

36740/ x = the percentage :)
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An equation was used to predict the number of possible enrollments in an afterschool program for the first 6 months of the year.
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55 + 80 + 95 + 100 +115 +90 =535 actual enrollments. 75 + 80 +85 +90 +95 + 100 = 525 predicted enrollments. 535 - 525 = 10 residuals.
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3 years ago
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