Answer:
By putting the coordinates on the graph I think
Step-by-step explanation:
There are no solutions to 2x-5y=9 but if you are solving it by y=mx+b form than your answer would look like this: y=2/5x-9/5
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)
<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
Answer:
2,21 * 10^20
Step-by-step explanation:
