Choose the correct conic section to fit the equationy=1/8x^2 a. Circle b. Ellipse c. Parabola d. Hyperbola

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:

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6 0

3 years ago

The weight of meteorite A is 5 times the weight of meteorite B. If the sum of their weights is 126 tons, find the weight of each

Sholpan [36]

Answer:

Step-by-step explanation:

if you divide 126 by 5, you get Meteorite B, 25.2, then minus that from 132, and you get Meteorite A, 106.8

To check, do 25.2 x 5 and you should get 132.

5 0

2 years ago

PLEASE HELP 7TH GRADE MATH

shepuryov [24]

21 is I

should be akakakdndfmdmsmwwkwksoizjdnsnwk

6 0

3 years ago

The regular price for a sweater is $48. The store is having buy one get one 1/2 off sale. If you buy 2 sweaters for that price w

IRISSAK [1]

25%

If you are buying 2 shirts for the same price, you have to calculate the cost of the shirt at half price first (1 piece of the pie) and then add it the original price (2 equal pieces of the pie since you just divided that to the price of the second shirt). That would be considered 3 out of 4 pieces off the pie or 75%. Your savings would be 25%.

4 0

3 years ago

Read 2 more answers

A survey in Men’s Health magazine reported that 39% of cardiologists said that they took vitamin E supplements. To see if this i

zzz [600]

Answer:

$z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615$

The p value for this case would be:

$p_v =2*P(z$

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39

Step-by-step explanation:

Information given

n=100 represent the random sample taken

X=36 represent the number of people that take E supplement

$\hat p=\frac{36}{100}=0.36$ estimated proportion of people who take R supplement

$p_o=0.39$ is the value that we want to test

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is equatl to 0.39 or not, the system of hypothesis are.:

Null hypothesis: $p=0.39$

Alternative hypothesis: $p \neq 0.39$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info we got:

$z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615$

The p value for this case would be:

$p_v =2*P(z$

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39

8 0

3 years ago