A sample proportion of 0.43 is found. To determine the margin of error for this statistic, a simulation of 100 trials is run, ea
2 answers:
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Answer:
a). Cost of 44 pupils = $14265
b). Least number of pupils = 31
Step-by-step explanation:
The given question is incomplete; here is the complete question.
The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(a) Find the cost when there are 44 pupils.
(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?
Let the equation representing the total cost of maintaining a school is,
C = ax + b
Where C = Total cost of maintaining a school
a = Fee per pupil
b = Fixed running cost
x = number of pupils
a). Cost of 50 pupils = $15705
Equation will be,
15705 = 50a + b -------(1)
Cost of 40 pupils = $13305
Equation will be,
13305 = 40a + b --------(2)
By subtracting equation (2) from equation (1),
15705 - 13305 = (50a + b) - (40a + b)
2400 = 10a
a = 240
From equation (1),
b = 3705
Equation representing the total cost will be,
C = 240x + 3705
If x = 44
C = 240(44) + 3705
C = $14265
b). If the fee per pupil 'a' = $360
Let the number of pupils = p
Total fee of 'p' pupils = $360p
Total cost to run the school will be = 3705 + 240p
For the school not to be in the loss,
360p ≥ 3705 + 240p
360p - 240p ≥ 3705
120p ≥ 3705
p ≥
p ≥ 30.875
Therefore, to run the school without loss, number pupils should be at least 31.
Answer:
The minimum number of words left to write is 215
Step-by-step explanation:
Let
w -----> the number of words left to write
we know that
The number of words left to write plus the words written so far must be greater than or equal to 500 words
Remember that the word "at least" means "greater than or equal to"
so
Solve for w
Subtract 285 both sides
The minimum number of words left to write is 215