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lana [24]
3 years ago
14

Urgent help please

Mathematics
1 answer:
andrew-mc [135]3 years ago
6 0

Mean is the arithmatic average of a given data set. So,

Mean= \frac{Sum of the data}{Number of the data}

There are seven data in the data set.

According to the given problem,

\frac{18+p+13+q+15+r+7 }{7} =11

\frac{53+p+q+r }{7} =11 Combine the like terms of the numerator.

\frac{53+p+q+r }{7}*7 =11 *7 Multiplying 7 to each sides of equation.

53 + p + q + r = 77

53 + p + q + r - 53 = 77 - 53 Subtract 53 from each sides.

p + q + r = 24

Next step is to divide each sides by 3 so that we can write this in form of mean of p, q and r. Therefore,

\frac{p + q + r}{3} =\frac{24}{3}

So, mean of p, q and r = 8.

I hope this helps you!.

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George has $49 which he decides to spend on x and y. commodity x costs $5 per unit and commodity y costs $11 per unit. he has th
FromTheMoon [43]

We are given the equations:

<span>5 x + 11 y = 49                    --> eqtn 1</span>

<span>u = 3 x^2 + 6 y^2               --> eqtn 2</span>

 

Rewrite eqtn 1 explicit to y:

11 y = 49 – 5 x

<span>y = (49 – 5x) / 11               --> eqtn 3</span>

 

Substitute eqtn 3 to eqtn 2:

u = 3 x^2 + 6 [(49 – 5x) / 11]^2

u = 3 x^2 + 6 [(2401 – 490 x + 25 x^2) / 121]

u = 3 x^2 + 14406/121 – 2940x/121 + 150x^2/121

u = 4.24 x^2 – 24.3 x + 119.06

Derive then set du/dx = 0 to get the maxima:

du/dx = 8.48 x – 24.3 = 0

solving for x:

8.48 x = 24.3

x = 2.87

 

so y is:

y = (49 – 5x) / 11 = (49 – 5*2.87) / 11

y = 3.15

 

Answer:

<span>George will choose some of each commodity but more y than x.</span>

7 0
3 years ago
Why is the fraction 6/10 not written 0.06?
Kaylis [27]
Because 10 divided by 6 equals 6 tenths not 6 hundredths
3 0
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Answer:

-x^{2} +2xy+10y^{2}

Step-by-step explanation:

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o-na [289]

Answer:

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The function for selling shirts is

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For there to be a profit, he must make more money selling the shirts than he spent on making the shirts

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15x > 50 + 7.5x

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To make a profit, Juan must sell at least 7 shirts

7 0
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Did I do this right? (Accounting)
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