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kow [346]
3 years ago
7

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

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The correct responses are;

(a) <u>The shape of the data remains the same</u>

(b) <u>The mean and median are increased by $1,000</u>

(c) <u>The standard deviation and interquartile range remain the same</u>

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

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The interquartile range, IQR = Q₃ - Q₁

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

Therefore;

  • <u>The interquartile range stays the same</u>

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