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2 years ago

7

2/5 of Janet's class prefers chocolate ice cream and 2/5 prefers vanilla ice cream. What part of the class prefers chocolate or

vanilla ice cream?
A. 0/5
B. 4/10
C. 2/5
D. 4/5

2 answers:

belka [17]2 years ago

6 0

The answer is D, 4/5, because 2/5+2/5=4/5.

AlexFokin [52]2 years ago

3 0

Answer:

The answer is D, 4/5, because 2/5+2/5=4/5.

Step-by-step explanation:

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Each year a car is worth 8% less than it was the year before. After 3 years, the car is worth $19,311.46. What was the original

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Answer:

$24,800

Step-by-step explanation:

If a car is worth 8% less after one year, then it is worth 92% of the original value.

original value = x

after 1 year: 0.92x

after 2 years: 0.92(0.92x)

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2 years ago

By how much does 5 exceed -4?<br>(a) 1<br>(b)-1<br>(c) 9<br>(d) -9

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Answer:

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3 years ago

Select all of the correct answers.

Korvikt [17]

Question:

Martin uses 5/8 of a gallon of paint to cover 4/5 of a wall. what is the unit rate at which Martian paints in walls per gallon

Answer:

0.78125 walls per gallon

<em></em>

Step-by-step explanation:

Given

$Paint = \frac{5}{8}\ gallon$

$Wall = \frac{4}{5}$

Required

Determine the unit rate

This is calculated as:

$Unit\ Rate = \frac{Wall}{Paint}$

Substitute values for wall and paint

$Unit\ Rate = \frac{5/8}{4/5}$

Simplify:

$Unit\ Rate = \frac{5}{8}/\frac{4}{5}$

Change to *

$Unit\ Rate = \frac{5}{8}*\frac{5}{4}$

$Unit\ Rate = \frac{5*5}{8*4}$

$Unit\ Rate = \frac{25}{32}$

$Unit\ Rate = 0.78125$

<em>Hence, the unit rate is: 0.78125 walls per gallon</em>

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2 years ago

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Hi! im in eighth grade too, and im learing the same stuff, i could probably help you!

7 0

3 years ago

Read 2 more answers

Let X represent the amount of energy a city uses (in megawatt-hours) in the Kanto region. Let Y represent the amount of mismanag

Setler [38]

Answer:

Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4:The value of E(y) is 4.6667.

Part 5:The value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6:The value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7:The value of E(x) is 3.6667.

Part 8:The value of E(x,y) is 36.

Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as $f_{xy}(x,y)\neq f_x(x).f_y(y)\\$

Step-by-step explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

$f(x,y)=\frac{1}{16}$ for $2\leq y\leq 2x\leq 10$

Now the distribution of x is given as

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy$

Here the limits for y are $2\leq y\leq 2x$ So the equation becomes

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy\\f_x(x)=\int\limits^{2x}_{2} \frac{1}{16} \, dy\\f_x(x)=\frac{1}{16} (2x-2)\\f_x(x)=\frac{x-1}{8} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5$

Part 2:

The probability is given as

$P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {\frac{x-1}{8}} \, dx +\int\limits^5_4 {\frac{x-1}{8}} \, dx\\P(X\leq 2 U X\geq 4)=\frac{1}{16}+\frac{7}{16}\\P(X\leq 2 U X\geq 4)=0.5$

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx$

Here the limits for x are $y/2\leq x\leq 5$ So the equation becomes

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx\\f_y(y)=\int\limits^{5}_{y/2} \frac{1}{16} \, dx\\f_y(y)=\frac{1}{16} (5-\frac{y}{2})\\f_y(y)=\frac{10-y}{32} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10$

So the value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4

The value is given as

$E(y)=\int\limits^{10}_2 {yf_y(y)} \, dy\\E(y)=\int\limits^{10}_2 {y\frac{10-y}{32}} \, dy\\E(y)=\frac{1}{32}\int\limits^{10}_2 {10y-y^2} \, dy\\E(y)=4.6667$

So the value of E(y) is 4.6667.

Part 5

This is given as

$f_{xy}(x)=\frac{f_{xy}(x,y)}{f_y(y)}\\f_{xy}(x)=\frac{\frac{1}{16}}{\frac{10-y}{32}}\\f_{xy}(x)=\frac{2}{10-y}$

So the value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6

The value is given as

$\geq M_{x,y}(y)=E(f_{xy}(x))=\int\limits^5_{y/2} {x f_{xy}(x)} \, dx \\M_{x,y}(y)=\int\limits^5_{y/2} {x \frac{2}{10-y}} \, dx \\M_{x,y}(y)=\frac{2}{10-y}\left[\frac{x^2}{2}\right]^5_{\frac{y}{2}}\\M_{x,y}(y)=\frac{2}{10-y}\left(\frac{25}{2}-\frac{y^2}{8}\right)\\M_{x,y}(y)=\frac{y+10}{4}$

So the value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7

The value is given as

$E(x)=\int\limits^{5}_1 {xf_x(x)} \, dx\\E(x)=\int\limits^{5}_1 {x\frac{x-1}{8}} \, dx\\E(x)=\frac{1}{8}\left(\frac{124}{3}-12\right)\\E(x)=\frac{11}{3} =3.6667$

So the value of E(x) is 3.6667.

Part 8

The value is given as

$E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xyf_{x,y}(x,y)} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xy\frac{1}{16}} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \frac{x}{16}\left[\frac{y^2}{2}\right]^{10}_2\, dx\\E(x,y)=\int\limits^{5}_1 3x\, dx\\\\E(x,y)=3\left[\frac{x^2}{2}\right]^5_1\\E(x,y)=36$

So the value of E(x,y) is 36

Part 9

The value is given as

$Cov(X,Y)=E(x,y)-E(x)E(y)\\Cov(X,Y)=36-(3.6667)(4.6667)\\Cov(X,Y)=18.8886\\$

So the value of Cov(x,y) is 18.8886

Part 10

The variables X and Y are considered independent when

$f_{xy}(x,y)=f_x(x).f_y(y)\\$

Here

$f_x(x).f_y(y)=\frac{x-1}{8}\frac{10-y}{32} \\$

And

$f_{xy}(x,y)=\frac{1}{16}$

As these two values are not equal, this indicates that X and Y are not independent variables.

4 0

3 years ago