(20 POINTS AND BRAINLIEST FOR BEST)

Jason has decided to sell one of his baseball cards. The card is valued at $51.00. The store is only offering Jason $29.00. How

sveta [45]

Well, just take the difference, 51-29, and 22 is how much less his card it being valued at. Or you can do 29-55, which would be -22, showing the amount and that it's valued less.

3 0

2 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

(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;

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

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

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

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:

brainly.com/question/9995782

6 0

2 years ago

Part B

abruzzese [7]

Answer:

The method regarding the sampling to pick 200 sales representative is to make 20 groups of 10.

What is sampling?

Sampling simply means a process that's used in statistics where a predetermined number of observations are selected from the larger population.

In this case, the best way for the sales director to include 200 sales representatives, with an equal number from each region is simply to make 20 groups of 10.

Step-by-step explanation: correct on plato:)

6 0

2 years ago

Can someone help me with this question, again?

givi [52]

A. 4*1 - 3*2 = 4 - 6 = -2 (X
B. 4*4 - 3*2 = 16 - 6 = 10 (O
C. 4*7 - 3*6 = 28 - 18 = 10 (O
D. 4*4 - 3*(-3) = 16 + 9 = 25 (X

so the answers are B and C

7 0

3 years ago

This is number 8. And I need help.

ANEK [815]

It would be the first one because it says that there are 4 times as many red roses as pink roses and so that would be represented r=4p

Hope this helps! Good luck!

6 0

3 years ago

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