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3 years ago

Plz help me only 5 mins left

Mathematics

2 answers:

blondinia [14]3 years ago

6 0

Answer:

3x7 • (4x2 - 5x + 3)

Step-by-step explanation:

End of step 1- (3 • (x7)) • ((22x2 - 5x) + 3)

End step 2- 3x7 • (4x2 - 5x + 3)

Step 3-Trying to factor by splitting the middle term

3.1 Factoring 4x2-5x+3

The first term is, 4x2 its coefficient is 4 .

The middle term is, -5x its coefficient is -5 .

The last term, "the constant", is +3

Step-1 : Multiply the coefficient of the first term by the constant 4 • 3 = 12

Step-2 : Find two factors of 12 whose sum equals the coefficient of the middle term, which is -5 .

-12 + -1 = -13

-6 + -2 = -8

-4 + -3 = -7

-3 + -4 = -7

-2 + -6 = -8

-1 + -12 = -13

1 + 12 = 13

2 + 6 = 8

3 + 4 = 7

4 + 3 = 7

6 + 2 = 8

12 + 1 = 13

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Final result :

3x7 • (4x2 - 5x + 3)

Elena L [17]3 years ago

6 0

I think the correct answer is B

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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:

brainly.com/question/9995782

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3 years ago

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shusha [124]

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3 years ago

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Combine these two equations together.

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Combine like terms.

-3x^3 - 3x^2 + 7x + 2

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