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2 years ago

92, 63, 22, 80, 63, 71, 44, 35 Mean:? Median:? Mode:? Range:?

Mathematics

2 answers:

Nutka1998 [239]2 years ago

4 0

Answer:

−

Step-by-step explanation:

subtract and add

Black_prince [1.1K]2 years ago

3 0

The mean is 58.75 . The median is 63. The mode is also 63 . And the range is 70

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Every economic decision has only positive benefits. gained opportunities. unavoidable limitations. a consequence or tradeoff.

Alina [70]

Every economic decision has "a consequence or tradeoff" - this final answer choice is correct. Every time that an individual, business, or institution makes an economic decision, they always forgo an opportunity to use the same capital or resources for other endeavors. As such, there is a tradeoff incurred by not making the decision to use the resource in another manner. This is known as opportunity cost and is one of the fundamental tenets of economic theory.

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4 years ago

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PLEASE HELP!!! Two trains simultaneously left points M and N and headed towards each other. The distance between point M and poi

iogann1982 [59]

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Step-by-step explanation:

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3 years ago

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Solve the system of equations:

Ahat [919]

Answer:

Step-by-step explanation:

z = 3

2y + z = 1

2y + 3 = 1 {substituting z =3}

2y = 1 -3

2y = -2

y = - 2/2

y = -1

2x +3y +2z = 13

2x + 3 * (-1) + 2*3 = 13 { {substituting z =3} and y = -1}

2x -3 + 6 =13

2x + 3 = 13

2x = 13 - 3

2x= 10

x = 10/2

x = 5

8 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Eliza estimated she would spend $120 at the grocery store. She actually spent $95. What is the percent error?

Amanda [17]

Answer: 26.32% (rounded to the hundredths)

6 0

3 years ago