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Tema [17]
3 years ago
14

Which of the following functions is not a linear function?

Mathematics
1 answer:
Bad White [126]3 years ago
5 0

Answer:

Hello,

not linear function :

f(x) = -2

f(x) = x²

f(x) = x - 2

Step-by-step explanation:

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Annual salary plus bonus data for chief executive officers are presented in the BusinessWeek Annual Pay Survey. A preliminary sa
Anuta_ua [19.1K]

Answer:

175

Step-by-step explanation:

The sample size n in<em> Simple Random Sampling</em> is given by

\bf n=\frac{z^2s^2}{e^2}

<em> where z = 1.96 is the z-score associated with a 95% confidence level ans s the standard deviation. </em>

Replacing values

\bf n=\frac{(1.96)^2(675)^2}{(100)^2}=175

So, there should be 175 chief executive officers in a sample if we want to estimate the population mean annual salary plus bonus with a margin of error of $100,000

7 0
3 years ago
Which of the following cannot be used to express the number 8? eight more than zero +8 eight below zero positive eight
MrRa [10]

Eight below zero would express the value -8, negative eight,

but not 8.

7 0
3 years ago
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2.In quadratic equation ax2 + bx + c = 0, if discriminant is D= b2 - 4ac, then roots of the quadratic equation are
tino4ka555 [31]

Answer:

(2) Real and equal (i.e., repeated roots), if D = 0.

Step-by-step explanation:

.In a quadratic equation ax2 + bx + c = 0, if discriminant is D= b2 - 4ac, then roots of the quadratic equation are

Real and equal (i.e., repeated roots), if D = 0.

If the D > b² - 4ac then it's real and distinct.

3 0
3 years ago
7. It is best to save _____ months of fixed expenses for an emergency fund. (1 point)
patriot [66]
I think the best answer is 3 to 9 but thats not a answer so 3 to 6
5 0
4 years ago
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The general solution of 2 y ln(x)y' = (y^2 + 4)/x is
Sav [38]

Replace y' with \dfrac{\mathrm dy}{\mathrm dx} to see that this ODE is separable:

2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}

Integrate both sides; on the left, set u=y^2+4 so that \mathrm du=2y\,\mathrm dy; on the right, set v=\ln x so that \mathrm dv=\dfrac{\mathrm dx}x. Then

\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v

\implies\ln|u|=\ln|v|+C

\implies\ln(y^2+4)=\ln|\ln x|+C

\implies y^2+4=e^{\ln|\ln x|+C}

\implies y^2=C|\ln x|-4

\implies y=\pm\sqrt{C|\ln x|-4}

4 0
3 years ago
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