Answer:
2/4
Step-by-step explanation:
Using a graph, you would start at (-1,2). Next you would move up twice. Then you would move 4 lines to the right. You'll land on (3,4)
Answer:
They lose about 2.79% in purchasing power.
Step-by-step explanation:
Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.
Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about
... (1 +0.03/365)^365 ≈ 1.03045326
Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...
... 1.03045326 / 1.06 ≈ 0.9721 . . . items
That is, the loss of purchasing power is about ...
... 1 - 0.9721 = 2.79%
_____
If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.
Answer:
The number of wrapping paper sold was 32 and the number of magazines sold was 40
Step-by-step explanation:
Let
x ----> the number of wrapping paper sold
y ----> the number of magazines sold
we know that
The classes sold 72 items
so
----> equation A
The classes earned $222 for their school
so
----> equation B
Solve the system of equations by graphing
Remember that the solution of the system of equations is the intersection point both graphs
using a graphing tool
The solution is (32,40)
see the attached figure
therefore
The number of wrapping paper sold was 32 and the number of magazines sold was 40
Circumference of circular path=pi*d^2/4
=44.156
length of light=1.45
light needed=44.156/1.45
=31 approx
