Value Grocery Mart and Market City are both having a sale on the same popular crackers. McKayla is trying to determine which sale is the better deal. Using the given table and equation, determine which store has the better deal on crackers? Explain your reasoning. (Remember to round your answers to the nearest penny.)

Value Grocery Mart:
Number of Boxes of Crackers
3
6
9
12
Cost (in dollars)
5
10
15
20

Answers

Answer 1

Answer: value grocery store

Step-by-step explanation: Because when you divide all the numbers by each other your gonna get 0.6 .

Answer 2

The same popular crackers are on sale at Value Grocery Mart and Market City for 5.25, 10.5, 15.75, and 21 cents.

What are the differences between chart and tabular forms?A chart is one that displays data using symbols such as bars in a bar chart, columns in a line chart, or slices in a pie chart. A chart can represent tabular numerical data, functions, or specific types of quality structures to convey a variety of information.Using a tabular form and a single page, users can change multiple rows and columns in a table at the same time.A chart displays data in the form of a table, graph, or diagram. It can convey vast amounts of information in a variety of ways.

As the cost per pack is C=1.75b. C is the Cost in dollars.

3 × 1.75 = 5.25

6 × 1.75 = 10.5

9 × 1.75 = 15.75

12 × 1.75 = 21

Here, the table is given below

The complete question is:

Value Grocery Mart and Market City are both having a sale on the same popular crackers. McKayla is trying to determine which sale is the better deal: Using the given table and equation, determine which store has the better deal on crackers? Explain your reasoning. (Remember to round your answers to the nearest penny.) Value Grocery Mart: 12 6 3 20 15 Number of Boxes of Crackers 10 5 Cost (in dollars) Market City: c = 1.75b, where c represents the cost in dollars, and b represents the number of boxes of crackers

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Value Grocery Mart And Market City Are Both Having A Sale On The Same Popular Crackers. McKayla Is Trying

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(-infinity, 10]

Step-by-step explanation:

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Answer:

y = - 3x - 2

Step-by-step explanation:

Step 1:

y = mx + b     Slope Intercept Form

Step 2:

y = - 3x + b     Input Slope

Step 3:

4 = - 3 ( - 2 ) + b     Input x and y

Step 4:

4 = 6 + b     Multiply

Step 5:

- 2 = b       Subtract 6 on both sides

Answer:

y = - 3x - 2

Hope This Helps :)

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Answer:It take her 20 minutes, but I dont know how to graph yet, so I can only tell you that xx=20

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Answers

Answer:

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Step-by-step explanation:

To solve this, you need to do fraction cross-multiplying.

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

Answer is C. 9 meters
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Answers

Answer:

x= - 8/9

Step-by-step explanation:

The random variables X and Y are jointly continuous, with a joint PDF of the form fX,Y(x,y)={cxy,0,if 0≤x≤y≤1,otherwise, where c is a normalizing constant. For x∈[0,0.5] , the conditional PDF fX|Y(x|0.5) is of the form axb . Find a and b .

Answers

Answer:

The value of a and b are 8 and 1 respectively

Step-by-step explanation:

[tex]f_{x,y}(x,y)=\left\{\begin{matrix}cxy ,\text{if} 0 \leq x \leq y \leq 1\\ 0 , \text{otherwise}\end{matrix}\right.[/tex]

[tex]\int_{0}^{1}\int_{x}^{1}cxy \text{dy dx}=1\\\\c\int_{0}^{1}x\left ( \frac{y^2}{2}\right )_{x}^{1} \text{ dx}=1\\\\\frac{c}{2}\int_{0}^{1}x\left ( 1-y^2\right )\text{dx}=1\\\\\frac{c}{2}\int_{0}^{1}(x-x^3)\text{dx}=1\\\\\frac{c}{2}\left [ \frac{x^2}{2}-\frac{x^5}{4} \right ]_{0}^1=1\\\\\frac{c}{2}\left [ \frac{1}{2}-\frac{1}{4} \right ]=1\\\\\frac{c}{2}\left ( \frac{1}{4} \right )=1\\\\c=8[/tex]

Conditional pdf of [tex]f_{x,y}(x|0.5)[/tex]

[tex]f_{x,y}(x|0.5)=\frac{f_{xy}(x,y=0.5)}{f_{y}y}[/tex]

Normalizing pdf of 1

[tex]f_{y}y=\int_{0}^{y} 8yx \text{dx}[/tex]

[tex]f_{y}y=8y[\frac{x^2}{2}]_{0}^{y}\\f_{y}y=4y(y^2)\\f_{y}y=4y^3\\f_{x|y}(x|0.5)=\frac{8x(0.5)}{4(0.5)^3}=\frac{2x}{0.25}=8x[/tex]

We are given that PDF [tex]f_{x|y}(x|0.5)[/tex] is of the form [tex]ax^b[/tex]

So, on comparing 8x with [tex]ax^b[/tex]

So, a = 8 , b = 1

So, the value of a and b are 8 and 1 respectively

The resulting value of a and b given  the conditional PDF of fX|Y(x|0.5) is mathematically given as

a = 8 , b = 1 .

What are the values of a and b?

Question Parameters:

a joint PDF of the form fX,Y(x,y)={cxy,0,if 0≤x≤y≤1,

For x∈[0,0.5]

the conditional PDF fX|Y(x|0.5)

Generally, the equation for the function of (x,y)   is mathematically given as

[tex]f_{x,y}(x,y)=\left\{\begin{matrix}cxy ,\text{if} 0 \leq x \leq y \leq 1\\ 0 ,\end{matrix}\right.[/tex]

Therefore

[tex]\int_{0}^{1}\int_{x}^{1}cxy \text{dy dx}=1[/tex]

[tex]c\int_{0}^{1}x\left ( \frac{y^2}{2}\right )_{x}^{1} \text{ dx}=1[/tex]

[tex]\frac{c}{2}\left [ \frac{x^2}{2}-\frac{x^5}{4} \right ]_{0}^1=1[/tex]

[tex]\\\\\frac{c}{2}\left ( \frac{1}{4} \right )=1[/tex]

C=8

Where, the conditional PDF fX|Y(x|0.5)

[tex]f_{x,y}(x|0.5)=\frac{f_{xy}(x,y=0.5)}{f_{y}y}[/tex]

We have

[tex]f_{y}y=8y[\frac{x^2}{2}]_{0}^{y}\[/tex]

[tex]f_{y}y=4y^3[/tex]

[tex]f_{x|y}(x|0.5)=\frac{8x(0.5)}{4(0.5)^3}[/tex]

f_{x|y}(x|0.5)=8x

In conclusion. with the pdf in the form of ax^2

8x comparing ax^b

The value of a = 8 , b = 1 .

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Ali"s dog weighs 8 times more than her cat. Together the animals weigh 54 lbs. How much does Ali's dog weigh?​

Answers

Answer:

48 lbs

Step-by-step explanation:

say that the cat weighs x pounds. then the dog weighs 8x pounds. together they weigh 54 pounds. So our equation is x+8x=54. solve for x to get the cat's weight, and you get x=6. The cat weighs 6 lbs. all I need to do now is multiply 6 by 8 so this the dog weighs 8 times as much as the cat. 8•6=48. the dog weighs 48 pounds.

Don’t understand this question

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i believe it would be A D E
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