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Trava [24]
3 years ago
13

Help me please!!

Mathematics
2 answers:
pogonyaev3 years ago
8 0

Answer:

\mid -2\frac{2}{3}\mid =2\frac{2}{3}

Step-by-step explanation:

We are given that a number in mixed fraction

-2\frac{2}{3}

We have to find its absolute value .

Absolute value of any number :If a  be number then its absolute value is given by

\mid a\mid

Absolute value of -2\frac{2}{3} is given by

\mid -2\frac{2}{3}\mid =2\frac{2}{3}

Absolute value of any number is the distance  or magnitude of that number from zero on the number line.

Distance always positive.

Hence, the absolute value

\mid -2\frac{2}{3}\mid =2\frac{2}{3}

cestrela7 [59]3 years ago
6 0

Absolute value is just the number's distance from zero, so I beleive you just drop the negative sign. It would be 2 2/3

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Jill traveled to Canada and bought a new lamp for $460 Canadian dollars. In U.S. dollars (to the nearest cent) this is equivalen
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Read 2 more answers
Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)
viktelen [127]
\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}

The first line segment can be parameterized by \mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle with 0\le t\le1. Denote this first segment by C_1. Then

\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt
=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt
=108

The second line segment (C_2) can be described by \mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle, again with 0\le t\le1. Then

\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt
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Finally,

\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3
5 0
3 years ago
A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec
VMariaS [17]

Answer:

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=19.1 represent the sample mean

\mu population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=9-1=8

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that t_{\alpha/2}=

Now we have everything in order to replace into formula (1):

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

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I assume that in this question, you are asked to evaluate the value(s) of r. To answer that question, dividing 1252 by 48 gives us 26.0833 or 26. Mathematically expressing the solution,
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