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2 years ago

What statement is NOT true comparing the two box and whisker plots?

Mathematics

2 answers:

ivolga24 [154]2 years ago

7 0

Answer: C

Step-by-step explanation:

The minimum is the line furthest to the left, and they are both lined up. The median, the line inside the box, is on 12. The third quartile, the edge of the box on the right hand side, are the same for both graphs. However, test 1 has a lower first quartile, the edge of the box on the right hand side, than test 2.

shutvik [7]2 years ago

3 0

C because it is higher den the first

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If A(0,0) and B(2,5) find AB

mojhsa [17]

Distance = sqrt[ (5-0)^2 + (2-0)^2]
Sqrt (5^2 + 2^2)
Sqrt(25+4)
Sqrt(29)

4 0

3 years ago

Teacher raises A school system employs teachers at

Cerrena [4.2K]

By adding a constant value to every salary amount, the measures of

central tendency are increased by the amount, while the measures of

dispersion, remains the same

The correct responses are;

(a) The shape of the data remains the same

(b) The mean and median are increased by $1,000

Reasons:

The given parameters are;

Present teachers salary = Between $38,000 and $70,000

Amount of raise given to every teacher = $1,000

Required:

Effect of the raise on the following characteristics of the data

(a) Effect on the shape of distribution

The outline shape of the distribution will the same but higher by $1,000

(b) The mean of the data is given as follows;

$\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}$

Therefore, following an increase of $1,000, we have;

$\overline x_{New} = \dfrac{\sum (f_i \cdot (x_i + 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i + f_i \cdot 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i}$

$\overline x_{New} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + 1000 = \overline x + 1000$

Therefore, the new mean, is equal to the initial mean increased by 1,000

Median;

Given that all salaries, $x_i$ , are increased by $1,000, the median salary, $x_{med}$ , is also increased by $1,000

Therefore;

The correct response is that the median is increased by $1,000

Where;

n = The number of teaches;

Given that, we have both a salary, $x_i$ , and the mean, $\overline x$ , increased by $1,000, we can write;

$\sigma_{new} =\sqrt{\dfrac{\sum \left ((x_i + 1000) -(\overline x + 1000)\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}} =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}} = \sigma$

Therefore;

$\sigma_{new} = \sigma$ ; The standard deviation stays the same

Interquartile range;

The interquartile range, IQR = Q₃ - Q₁

New interquartile range, IQR $_{new}$ = (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR

Therefore;

The interquartile range stays the same

Learn more here:

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6 0

3 years ago

3. Complete the square for 3x2 - 6x = 21. Help

PSYCHO15rus [73]

Answer:

x=1+2√2 or x=1−2√2

Step-by-step explanation:

Let's solve your equation step-by-step.

3x2−6x=21

Step 1: Since the coefficient of 3x^2 is 3, divide both sides by 3.

3x2−6x

x2−2x=7

Step 2: The coefficient of -2x is -2. Let b=-2.

Then we need to add (b/2)^2=1 to both sides to complete the square.

Add 1 to both sides.

x2−2x+1=7+1

x2−2x+1=8

Step 3: Factor left side.

(x−1)2=8

Step 4: Take square root.

x−1=±√8

Step 5: Add 1 to both sides.

x−1+1=1±√8

x=1±√8

x=1+2√2 or x=1−2√2

7 0

3 years ago

The area of a squared lawn and rectangular vegetable garden are equal.The perimeter of lawn is 24 m and its side is twice the br

rjkz [21]

Answer:

g=12

Step-by-step explanation:

4(2y)=24

8y=24

$y = \frac{24}{8}$

y=3

${4y}^{2} = yg$

Since we know the value of y we substitude

${4 \times 3}^{2} = 3g$

36=3g

g=12

6 0

2 years ago

A fitness instructor tested the arm strength of members of a fitness dass to compare their scores to the national average. The t

tester [92]

Answer:

A) Median: 7

Mode: 8

Range: 5

B) Tyler is wrong, the greater the MAD value, the greater the variability

Step-by-step explanation:

A) Given the data set: 4,6,6,7,7,8,8,8,9

The median is 7 (the middle point in the ordered list)

The mode is 8 (the most repeated number)

The range is 5 (the difference between the highest and lowest values)

3 0

3 years ago