It is question 13 if you are wondering

Which graph represents the following system of inequalities? Y>5x-1 and then y less than or equal to x+3 please it's for Plat

Stolb23 [73]

Answer:

y=3

Step-by-step explanation:

If we use LCM and take 3 plus y it equals x so if x equals x it is then y

6 0

3 years ago

How do you solve 4x^2+20x+25=0 by completing the square

svlad2 [7]

(2x+5)(2x+5)= 0

2x+5=0. 2x+5=0

x= -5/2, x= -5/2

3 0

3 years ago

32 divided by 8 what is the answer?

ValentinkaMS [17]

The answer is The number 4

6 0

3 years ago

Read 2 more answers

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

(c) The standard deviation and interquartile range remain the same

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

(c) The standard deviation, σ, is given by $\sigma =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}}$ ;

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;

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

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

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