Ms. Thornton is buying treats for her top 10 algebra students in each of her two classes (20 treats total). She’s only willing t
1 answer:
Answer:
Mrs Thornton can buy almost 17 candy packs if she has $30 to spend and one pack cost $1.75.
No, it will not be enough because total students are 20 and total candy packs that can be bought are 17.
Step-by-step explanation:
Total treats to buy = 20
Total cost to spend = $30
Price of one candy pack = 1.75
We need to find:
a) What is the maximum number of packs that she will be able to buy?
Price of 1 candy pack = $1.75
If she spends $30, divide 30 by 1.75 to get the number of candy packs she can buy
Number of candy packs =
So, Mrs Thornton can buy almost 17 candy packs if she has $30 to spend and one pack cost $1.75
b) Will that be enough for all of her students?
No, it will not be enough because total students are 20 and total candy packs that can be bought are 17.
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Answer:
- 0 ≤ m ≤ 7
- 0.4541 cm/month; average rate of growth over last 4 months of study
Step-by-step explanation:
<u>Part A</u>:
The study was concluded after 7 months. The fish cannot be expected to maintain exponential growth for any significant period beyond the observation period. A reasonable domain is ...
0 ≤ m ≤ 7
__
<u>Part B</u>:
The y-intercept is the value when m=0. It is the length of the fish at the start of the study.
__
<u>Part C</u>:
The average rate of change on the interval [3, 7] is given by ...
(f(7) -f(3))/(7 -3) = (4(1.08^7) -4(1.08^3))/4 = 1.08^3·(1.08^4 -1)
≈ 0.4541 cm/month
This is the average growth rate of the fish in cm per month over the period from 3 months to 7 months.
