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Fiesta28 [93]
3 years ago
10

Help!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer: the answer is The best measure of the center for this data is the median which has a value of 23

Step-by-step explanation: Hope this help plz mark me as brainiest

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What is the equation for the values in the table
devlian [24]

Step-by-step explanation:

Taking any two points through which the line passes;(-10,-3) and (-5,-1)

The equation of the line is given by

\frac{y2 - y1}{x2 - x1}  =  \frac{y - y1}{x - x1}

=>  \frac{ - 1 - ( - 3)}{ - 5 - ( - 10)}  =  \frac{y - ( - 3)}{x -  ( - 10)}  =  \frac{y + 3}{x + 10}

=>  \frac{ - 1 + 3}{ - 5 + 10}  =  \frac{y + 3}{x + 10}

=>  \frac{2}{5}  =  \frac{y + 3}{x + 10}

=> 5(y + 3) = 2(x + 10)

=> 5y+15 = 2x+20

=> 5y =2x+5

=> y= \frac{2}{5} x + 1

option C

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3 years ago
Susan's weekly earnings were proportional to the number of hours she worked. This table shows the number of hours Susan worked a
oksano4ka [1.4K]

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3 years ago
Let f(x) = e ^3x/5x − 2. Find f'(0).
Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

Our friend asking what the actual function is has a point. I completed this under the assumption that what we have is:

f(x)=\frac{e^{3x}}{5x-2} and used the quotient rule to find the derivative, as follows:

f'(x)=\frac{e^{3x}(5)-[(5x-2)(3e^{3x})]}{(5x-2)^2} and simplifying a bit:

f'(x)=\frac{5e^{3x}-[15xe^{3x}-6e^{3x}]}{(5x-2)^2}and a bit more to:

f'(x)=\frac{5e^{3x}-15xe^{3x}+6e^{3x}}{(5x-2)^2} and combining like terms:

f'(x)=\frac{11e^{3x}-15xe^{3x}}{(5x-2)^2} and factor out the GFC in the numerator to get:

f'(x)=\frac{e^{3x}(11-15x)}{(5x-2)^2}  That's the derivative simplified. If we want f'(0), we sub in 0's for the x's in there and get the value of the derivative at x = 0:

f'(0)=\frac{e^0(11-15(0))}{(5(0)-2)^2} which simplifies to

f'(0)=\frac{11}{4} which translates to

The slope of the function is 11/4 at the point (0, -1/2)

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2 years ago
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