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Jobisdone [24]
3 years ago
14

Find the area of a rectangle that has the length of 2x-3 and the width of x .Someone painted an interior area of the rectangle a

nd it has a length of 3 and a width of x-2 .Find the size of the area that was not painted
Help me please

Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0

Answer:

The area that was not painted is (2x^{2}-6x+6)\ units^{2}

Step-by-step explanation:

step 1

Find the area of the rectangle

we know that

The area of a rectangle is equal to

A=LW

In this problem we have

L=2x-3

W=x

substitute

A=(2x-3)x\\A=(2x^{2}-3x)\ units^{2}

step 2

Find the area that was painted

A=(3)(x-2)\\A=(3x-6)\ units^{2}

step 3

Find the area that was not painted

Subtract the area that was painted from the total area of rectangle

so

A=(2x^{2}-3x)-(3x-6)=(2x^{2}-6x+6)\ units^{2}

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Answer:

From the estimation,

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Step-by-step explanation:

Given the conversion rate between the pounds and the euro currencies, we are to convert an amount in pounds to euros

From the estimate;

3 pounds = 4 euros

Thus;

10.7 pounds = x euros

We are using x since the value in euros is not known presently.

To move from here, we simply cross multiply.

That would give;

3 * x = 4 * 10.7

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The radius, r, of a circle is one-half of the length of its diameter, d. r= r= d + r = d 2 r = d r = d
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If the radius is HALF of the diameter, then the expression would be 1/2r=d or d/2=r.

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Which could be a conditional relative frequency table? A 4-column table with 3 rows. The first column has no label with entries
kotykmax [81]

Answer:

<u>II. Second table</u>

              A              B            Total

C           0.25         0.75          1.00

D           0.35         0.65          1.00

Total     0.30         0.70          1.00

Explanation:

<h2>Tables</h2>

<u>I. First table </u>

              A              B            Total

C           0.25          0.25          0.50

D           0.25          0.25          0.50

Total     0.50          0.25           1.00

<u>II. Second table</u>

              A              B            Total

C           0.25         0.75          1.00

D           0.35         0.65          1.00

Total     0.30         0.70          1.00

<u>III. Third table</u>

<u></u>

             A              B            Total

C          0.75         0.25          0.50

D          0.25         0.75          0.50

Total    0.50         0.50          1.00

<u>IV. Fourth table</u>

              A              B            Total

C          0.65         0.35         1.00

D          0.35         0.65          1.00

Total     1.00          1.00          1.00

<h2>Solution</h2>

A <em>conditional relative frequency table</em> shows the relative frequencies determined upon a row or column.

There are two types of relative conditional frequency table: 1) row conditional relative frequency, and 2) column conditional relative frequency.

When you divide the joint frequency by the marginal frequency of the column total you have the row conditional frequency table. When you dividethe joint frequency by the row total you have the colum conditional frequency table.

In a row conditional relative frequency each total of the right hand column equals 1. This is the case of the second table.

In a column conditional relative frequency each total of the bottom row equals 1. This is not happening with any of the shown tables.

Hence, only the second table could be a conditional relative frequency table.

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