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alexandr402 [8]
3 years ago
13

Which graph would you want to use if you worked for Brand A and wanted to show that your product is preferred almost as often as

Brand B?
A. Neither graph is preferred

B. Either graph could be used.

C. The graph on the left.

D. The graph on the right.

Mathematics
2 answers:
jasenka [17]3 years ago
5 0
It would be C. The right one makes it look horrible compared to the other, if someone did not look at the graph.
timurjin [86]3 years ago
5 0

Answer:

Option C is right

Step-by-step explanation:

Given is a graph showing the preferences of Brands A and B.

We have to select the one out of two which shows that your product is preferred almost as often as Brand B

This implies that the preferences of both should be almost equal.

In other words, heights of these products in the graph of rectangular boxes must be almost the same

We find that right graph has height difference very much high.

But graph on left has heights almost close to each other

So The graph on the left i.e. option C is right answer

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Step-by-step explanation:

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Identify p, b, and a. Do not solve for the unknown.<br><br> 10.9 is what percent of 34?
Rashid [163]
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What are the solutions?<br>8x^2-12x-23=-5​
erastova [34]

Answer:

x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

Step-by-step explanation:

Solve for x over the real numbers:

8 x^2 - 12 x - 23 = -5

Divide both sides by 8:

x^2 - (3 x)/2 - 23/8 = -5/8

Add 23/8 to both sides:

x^2 - (3 x)/2 = 9/4

Add 9/16 to both sides:

x^2 - (3 x)/2 + 9/16 = 45/16

Write the left hand side as a square:

(x - 3/4)^2 = 45/16

Take the square root of both sides:

x - 3/4 = (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

x = 3/4 + (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

Answer:  x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

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Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)​
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xe^y+4\ln y=x^2

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\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

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e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

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If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

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