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maksim [4K]
2 years ago
11

ZC and ZD are vertical angles with mZC = -3x+58 and mZD= x - 2.

Mathematics
1 answer:
kogti [31]2 years ago
3 0

Answer:

×=-2

0=×-2

×=-2

33333333

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I need help on this lol
Snezhnost [94]
Terms are 3x, 5, 2x, and x.
Like terms are 3x and 2x.
Coefficients are 3 and 2.
Constant is 5.

lmk if anything is incorrect!!
6 0
3 years ago
Read 2 more answers
Instruction
slavikrds [6]
<h3>Answer: Choice C.  (-x, -y)</h3>

Explanation:

Focus on one point, such as A = (1,2). Note how it moves to point A ' = (-1, -2). Both x and y coordinates have been flipped from positive to negative. The rule therefore is (x,y) \to (-x,-y). This describes a 180 degree rotation (either clockwise or counterclockwise, it doesn't matter). Points B and C follow the same idea.

Side note: Lines AA', BB' and CC' all go through the origin (0,0).

8 0
3 years ago
2
antiseptic1488 [7]

Answer: 123

Step-by-step explanation: Stop cheating idiot

5 0
3 years ago
A rectangle has a length of 6.5m and a width of 7.3m what is the area and the perimeter
Bad White [126]

Step-by-step explanation:

<h3><em><u>Given</u></em><em><u>:</u></em></h3>

Length of the rectangle = 6.5 m

Width of the rectangule = 7.3 m

<h3><em><u>Then</u></em><em><u>:</u></em></h3>

<u>First</u><u> </u><u>case</u><u>,</u>

Area of the rectangle

= length × width

= 6.5 m × 7.3 m

= <em><u>47.45</u></em><em><u> </u></em><em><u>s</u></em><em><u>q</u></em><em><u>.</u></em><em><u>m</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em><em><u>(</u></em><em><u>i</u></em><em><u>)</u></em>

<u>Second</u><u> </u><u>case</u><u>,</u>

Perimeter of the rectangle

= 2(length + width)

= 2(6.5 + 7.3)m

= 2 × 13.8 m

= <em><u>27</u></em><em><u>.</u></em><em><u>6</u></em><em><u> </u></em><em><u>m</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em><em><u>(</u></em><em><u>ii</u></em><em><u>)</u></em>

5 0
2 years ago
Read 2 more answers
Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{1048}{5}}

Standard deviation = \sqrt{209.6}

Standard deviation = 14.48

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{14.48}{142}*100

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0
3 years ago
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