Marissa wants to babysit to earn some extra money. She already has $250 in her bank account. The equation f(x) = 22x + 250 can b
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Step-by-step explanation:
Well in general, we can represent subtraction as:
"z" represents the difference, and it really just represents x with y taken away. So if we were to "give back" this y value, we should get "x".
This means that:
So one way to check, is adding the value that's being subtracted (y value) and the difference (z value), this should get you the value that is being subtracted from (x value). If you don't get the original value that's being subtracted from (x-value) then you know the answer you got is wrong.
The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data
Required values;
(a) The performance for the week for Park Street
- Revenue is <u>Q₂ < $7,500 < Q₃</u>
- The sales for the week is better than <u>72.91%</u> of all sales
The performance for the week for Bridge Road
- Revenue; <u>Q₂ < $7,100 < Q₃</u>
- The sale for the week is better than <u>59.87%</u> of all sales
(b) The mean is <u>$3611</u>
The median is $<u>3,600</u>
The standard deviation is $<u>3250</u>
The Interquartile range is $<u>6075</u>
Reason:
The table of values that maybe used to find a solution to the question is given as follows;
(a) Park Street revenue = $7,500
Bridge Road's revenue = $7,100
The two stores sold close to but below the 75th percentile
Bridge Road revenue;
The z-score is given as follows;
From the Z-Table, we have;
The percentile= 0.7291
- Therefore, the sale for the week for Park Street is better than <u>72.91%</u> of all the sales
Park Street revenue;
The z-score is given as follows;
From the Z-Table, we have;
The percentile = <u>0.5987</u>
- Therefore, the sale for the week is better than <u>59.87 %</u> of all the sales
(b) Given that the operating cost is $3,000, frim which we have;
The subtracted value is subtracted from the mean and median to find the new value
Profit = The revenue - Cost
New mean = 6611 - 3000 = 3611
- The new mean = <u>$3,611</u>
The new median = 6600 - 3000 = 3600
- The new median = <u>$3,600</u>
The standard deviation and the interquartile range remain the same, therefore, we have;
- The standard deviation = <u>$3,580</u>
The interquartile range = 9675 - 3600 = 6075
- The interquartile range = <u>6075</u>
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