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Len [333]
3 years ago
5

Refer to the following distribution of commissions:______. Monthly Commissions Class Frequencies $600 up to $800 3 $800 up to 1,

000 7 $1,000 up to $1,200 11 $1,200 up to $1,400 22 $1,400 up to $1,600 40 $1,600 up to $1,800 24 $1,800 up to $2,000 9 $2,000 up to $2,200 4 For the preceding distribution, what is the midpoint of the class with the greatest frequency?
Mathematics
1 answer:
blsea [12.9K]3 years ago
3 0

Answer:

 the midpoint of the class with the greatest frequency 1500

Step-by-step explanation:

The illustration of the dataset given can be well represented in a table format as shown below.

Class                      Frequency                     Relative Frequency

$600 - $800                   3                              \mathbf{   \dfrac{3}{120} = 0.025}

$800 - $1000                  7                              \mathbf{   \dfrac{7}{120} = 0.059}

$1000 - $1200                11                              \mathbf{   \dfrac{11}{120} = 0.092}

$1200 - $1400                22                             \mathbf{   \dfrac{22}{120} = 0.183}

$1400 - $1600                40                            \mathbf{   \dfrac{40}{120} = 0.333}

$1600 - $1800                 24                            \mathbf{   \dfrac{24}{120} = 0.2}

$1800 - $2000                 9                            \mathbf{   \dfrac{9}{120} = 0.075}

$2000 - $2200                4                             \mathbf{   \dfrac{4}{120} = 0.033}

Total                               120                                           1

Therefore, the midpoint of the class with the greatest frequency is between  $1400 - $1600  since the frequency 40 happens to be the greatest frequency

The midpoint =\dfrac{upper \ limit + \ lower \ limit}{2}

The midpoint =\dfrac{1400+1600}{2}

= 1500

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\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

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Step-by-step explanation:

Given the simultaneous equations

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\mathrm{Solve\:}\:x\left(2x+5\right)=3

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\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{Solve\:with\:the\:quadratic\:formula}

\mathrm{Quadratic\:Equation\:Formula:}

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

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