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vovikov84 [41]
3 years ago
5

A rectangle with vertices located at (1,2), (1,5), (3,5), and (3,2) is stretched horizontally by a factor of 3 with respect to t

he y-axis. What is the area of the image that is produced
Mathematics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

  18 square units

Step-by-step explanation:

The given rectangle has an x-dimension of 3-1 = 2, and a y-dimension of 5-2 = 3. Stretching the x-dimension by a factor of 3 will multiply it to ...

  2×3 = 6

and give you a rectangle 6 units wide by 3 units high. Its area will be ...

  (6 units)(3 units) = 18 units²

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A geologist examines 14 geological samples for iron concentration. The mean iron concentration for the sample data is 0.181 cc/c
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Answer:

The 90% confidence interval for the population mean iron concentration is between 0.167 cc/m³ and 0.195 cc/m³.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645*\frac{0.0318}{\sqrt{14}} = 0.0140

The lower end of the interval is the sample mean subtracted by M. So it is 0.181 - 0.0140 = 0.167 cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.181 + 0.0140 = 0.195 cc/m³.

The 90% confidence interval for the population mean iron concentration is between 0.167 cc/m³ and 0.195 cc/m³.

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Cual es la respuesta de (x10+10y12) al cuadrado​
padilas [110]

Answer:

\mathrm{Expand}\:\left(x^{10}+10y^{12}\right)^2:\quad x^{20}+20x^{10}y^{12}+100y^{24}

Step-by-step explanation:

\left(x^{10}+10y^{12}\right)^2

\gray{\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2}

\gray{a=x^{10},\:\:b=10y^{12}}

=\left(x^{10}\right)^2+2x^{10}\cdot \:10y^{12}+\left(10y^{12}\right)^2

\black{\mathrm{Simplify}\:\left(x^{10}\right)^2+2x^{10}\cdot \:10y^{12}+\left(10y^{12}\right)^2:}

\left(x^{10}\right)^2+2x^{10}\cdot \:10y^{12}+\left(10y^{12}\right)^2

\gray{\left(x^{10}\right)^2=x^{20}}

\gray{2x^{10}\cdot \:10y^{12}=20x^{10}y^{12}}

\gray{\left(10y^{12}\right)^2=100y^{24}}

=x^{20}+20x^{10}y^{12}+100y^{24}

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