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lbvjy [14]
3 years ago
13

PLEASE HELP 30 PTS!!! Select ALL the correct answers. Which expressions are equivalent to the following?

Mathematics
1 answer:
Oksanka [162]3 years ago
7 0

Answer:

  D:  -5(-6x^2 + x + 2)

  E:  5(2x + 1)(3x − 2)

Step-by-step explanation:

The <em>first two answer choices have incorrect constants</em> (25 and 3 vs -10). A factor of 5 is removed from the remaining answer choices, so let's remove a factor of 5 and see what we get:

  30x^2 -5x -10 = 5(6x^2 -x -2)

An additional x cannot be factored from the expression, so <em>choice C can be eliminated</em>.

Multiplying each of these factors by -1 will make the product correspond to answer choice D.

Factoring will make it correspond to answer choice E, best verified by finding the x-term of the product of the binomial factors:

  E:  2x(-2) +1(3x) = -x, as required

  F:  2x(2) -1(3x) = x, wrong sign

The equivalent expressions are those of choices D and E.

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C=  \left[\begin{array}{cc}1&-2\\-3&7\end{array}\right]  \\ C^{-1}= \left[\begin{array}{cc}7&2\\3&1\end{array}\right]  \\ \left[\begin{array}{cc}7&2\\3&1\end{array}\right] \times   \left[\begin{array}{ccccc}7&-28&-25&-35&-2\\-21&107&90&123&17\end{array}\right]  \\ =\left[\begin{array}{ccccc}49-42&-196+214&-175+180&-245+246&-14+34\\21-21&-84+107&-75+90&-105+123&-6+17\end{array}\right] \\ =\left[\begin{array}{ccccc}7&18&5&1&20\\0&23&15&18&11\end{array}\right]
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Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally
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Answer:

a) 0.1587 (or 15.87%)

b) 0.0478 (or 4.78%)

c) 0.7935 (or 79.35%)

Step-by-step explanation:

Find the following probabilities:

(a) the thickness is less than 3.0 mm

We need to find the area under the Normal curve with a mean of 4.5 mm and a standard deviation of 1.5 mm to the left of 3 mm

<h3>(See picture 1 attached) </h3>

We can do that with the help of a calculator or a spreadsheet.

<em>In Excel use </em>

<em>NORMDIST(3,4.5,1.5,1) </em>

<em>In OpenOffice Calc use  </em>

<em>NORMDIST(3;4.5;1.5;1) </em>

and we get the value 0.1587 (or 15.87%)

(b) the thickness is more than 7.0 mm

Now we need the area to the right of 7 (1 - the area to the left of 7)

<h3>(See picture 2 attached) </h3>

<em>In Excel use </em>

<em>1-NORMDIST(7,4.5,1.5,1)  </em>

<em>In OpenOffice Calc use  </em>

<em>1-NORMDIST(7;4.5;1.5;1) </em>

and we get the value 0.0478 (or 4.78%)

(c) the thickness is between 3.0 mm and 7.0 mm

We are looking for the area between 3 and 7

<h3>(See Picture 3) </h3>

Since the area under the Normal equals 1, we have

Area to the left of 3 + Area between 3 and 7 + Area to the right of 7 = 1

Hence,  

0.1587 +  Area between 3 and 7 + 0.0478 = 1

and

Area between 3 and 7 = 1 - 0.1587 - 0.0478 = 0.7935 (or 79.35%)

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