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Anna11 [10]
2 years ago
15

Kayla and latasha work for a department store the amount of sales they made over the last year was collected monthly in the data

was made into a box and whisker plot for each of them
a) what is the lowest amount of sales Latasha made in one month?

b) 50% of the sales Kayla made was between what two dollar amounts?

c)Compare the spread of Kayla’s middle 50% of sales to Latasha middle 50% of sales. which person has a 50% off sales was but I’ll provide math support your answer answer?

Mathematics
1 answer:
tresset_1 [31]2 years ago
4 0

Based on the information represented by the boxplot ;

  • Latasha's lowest sale amount = 50

  • Kayla's median is between 200 and 300

  • Latasha has a greater spread due to higher IQR value

1.) <em><u>The Lowest amount of sale made by Latasha in one month </u></em>

  • The minimum value is denoted by the starting position of the lower whisker on a boxplot.

  • Lowest amount of sale made by Latasha = 50

2.) <em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>made</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>Kayla</u></em><em><u> </u></em><em><u>:</u></em>

  • 50% of sales made marks the median value in a boxplot, it is denoted by the vertical line in between the box.

  • 50% of sales made by Kayla is between 200 and 300

  • With median sale value being 250

3.) <em><u>Spread</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>middle</u></em><em><u> </u></em><em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>:</u></em>

  • The measure of spread of the middle 50% of a distribution on a boxplot is the Interquartile range (IQR) of the distribution

  • IQR = Upper Quartile (Q3) - Lower quartile(Q1)

<u>For Latasha</u> :

  • Q3 = 450 (Endpoint of the box)
  • Q1 = 150 (starting point of the box)

  • IQR = 450 - 150 = 300

<u>For</u><u> </u><u>Kayla</u><u> </u><u>:</u><u> </u>

  • Q3 = 375 (Endpoint of the box)
  • Q1 = 100 (starting point of the box)
  • IQR = 375 - 100 = 275

  • Since, Latasha's IQR is greater than Kayla's, then Latasha has a greater mid 50% spread than Kayla.

Learn more :brainly.com/question/24582786

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For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
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Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

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