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alekssr [168]
3 years ago
9

Find the equation of the linear function represented by the table below in slope-

Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0

Answer:

y = -2x-5

Step-by-step explanation:

So to calculate the slope the equation is y1-y2/x1-x2. We substitute in two points, let's use (-3,1) as x1 and y1, and (1,-7) as x2 and y2. Put those points in the equation and you get -7-1/1-(-3), which is -2. Now you have y=-2x+b

To get b plug in a pair of points to the equations, say (1,-7). -7=-2(1)+b, solve that and you get b=-5.

So y=-2x-5.

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((60 * c) + 3300 ≤ 24000

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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
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Answer:

\frac{dy}{dx}=\frac{x(1-2lnx)}{x^{4}}

Step-by-step explanation:

To solve the question we refresh our knowledge of the quotient rule.

For a function f(x) express as a ratio of another functions u(x) and v(x) i.e

f(x)=\frac{u(x)}{v(x)}\\, the derivative is express as

\frac{df(x)}{dx}=\frac{v(x)\frac{du(x)}{dx}-u(x)\frac{dv(x)}{dx}}{v(x)^{2} }

from y=lnx/x^{2}

we assign u(x)=lnx and v(x)=x^2

and the derivatives

\frac{du(x)}{dx}=\frac{1}{x}\\\frac{dv(x)}{dx}=2x\\.

Note the expression used in determining the derivative of the logarithm function.it was obtain from the general expression of logarithm derivative i.e y=lnx\\\frac{dy}{dx}=\frac{1}{x}

If we substitute values into the quotient expression we arrive at

\frac{dy}{dx}=\frac{(x^{2}*\frac{1}{x})-(2x*lnx)}{x^{4}}\\\frac{dy}{dx}=\frac{x-2xlnx}{x^{4}}\\\frac{dy}{dx}=\frac{x(1-2lnx)}{x^{4}}

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A research study investigated differences between male and female students. Based on the study results, we can assume the popula
oksian1 [2.3K]

Answer: 0.0548

Step-by-step explanation:

Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.

Let \overline{X} represents the sample mean GPA for each student.

Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:

P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z

hence, the required probability is 0.0548.

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3 years ago
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