Express 44 copies in 4 minutes as a unit rate.
1 answer:
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Answer:
c(p) = {0.2 p ⇒ p < 1000
0.24 p - 40 ⇒ 1000 ≤ p ≤ 2000
0.29 p - 140 ⇒ p > 2000}
Step-by-step explanation:
* Lets explain how to solve the problem
- The profit is represented by p
1. If the profit is under $1,000, the commission rate is 20%
∵ The profit is p < 1000
∵ 20% of p = = 0.2 p
∵ c(p) is the function of the commission
∴ c(p) = 0.2 p when p < 1000
2. If the profit is at least $1,000 and less than or equal to $2,000, the
commission rate is 20% of the first $1,000 and 24% of the remainder
of the profit
- At least means greater than or equal
∵ The profit 1000 ≤ p ≤ 2000
- The commission is divided into 20% of first $1000 and 24% of
the reminder
∵ 20% of 1000 = = 200
∵ The remainder of the profit = p - 1000
∵ 24% of the remainder profit =
= 0.24(p - 1000) = 0.24 p - 240
∴ The total commission = 200 + 0.24 p - 240
∴ The total commission = 0.24 p - 40
∴ c(p) = 0.24 p - 40 when 1000 ≤ p ≤ 2000
3. If the profit is above $2,000, the rate is 20% of the first $1,000
of profit, 24% of the next $1,000 of profit, and 29% of the amount
of profit over $2,000
∵ The profit p > 2000
- The commission is divided into 20% of first $1000 and 24% of the
next $1,000 of profit, and 29% of the amount of profit over $2,000
∵ 20% of 1000 = = 200
∵ 24% of 1000 = = 240
- The amount of profit over $2,000 = p - 2000
∵ 29% of the amount of profit over $2,000 =
= 0.29(p - 2000)
= 0.29 p - 580
∴ The total commission = 200 + 240 + 0.29 p - 580
∴ The total commission = 0.29 p - 140
∴ c(p) = 0.29 p - 140 when p > 2000
* The commission function is:
c(p) = {0.2 p ⇒ p < 1000
0.24 p - 40 ⇒ 1000 ≤ p ≤ 2000
0.29 p - 140 ⇒ p > 2000}
The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.
Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars (not all zero) such that
which leads to the system of linear equations,
From the third equation, we have , and substituting this into the second equation gives
and in turn, . Substituting these into the first equation gives
which tells us that any value of will work. If
, then
and
. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.
Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.