Mia is working two summer jobs, making $11 per hour babysitting and making $20 per hour lifeguarding. In a given week, she can w
2 answers:
9514 1404 393
Answer:
- x + y ≤ 15
- 11x +20y ≥ 220
- 3 hours babysitting and 11 hours lifeguarding
Step-by-step explanation:
The two inequalities represent the two relations described in the problem statement.
x + y ≤ 15 . . . . . . . Mia works a maximum of 15 hours
11x + 20y ≥ 220 . . . . Mia makes at least $220
These are graphed in the attachment. The solution area is the doubly-shaded area with vertices at (9, 6), (0, 11) and (0, 15). One possible solution is shown at (3, 11), which represents ...
3 hours babysitting
11 hours lifeguarding . . . . . . . . total 14 hours for $253
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The two dot plots are missing, so i have attached it.
Answer:
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
Step-by-step explanation:
From the attached image, we are told to compare the means for each plot to the nearest tenth.
Mean = Σx/n
Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;
(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133
Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles
Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143
Mean of miles run at the end of the school year = 143/14 = 10.2 miles
Thus;
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
