Answer:
Step-by-step explanation:
You are being asked to compare the value of a growing infinite geometric series to a fixed constant. Such a series will always eventually have a sum that exceeds any given fixed constant.
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<h3>a)</h3>
Angelina will get more money from the Choice 1 method of payment. The sequence of payments is a (growing) geometric sequence, so the payments and their sum will eventually exceed the alternative.
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<h3>c)</h3>
For a first term of 1 and a common ratio of 2, the sum of n terms of the geometric series is given by ...
Sn = a1×(r^n -1)/(r -1) . . . . . . . . . . series with first term a1, common ratio r
We want to find n such that ...
Sn ≥ 1,000,000
1 × (2^n -1)/(2 -1) ≥ 1,000,000
2^n ≥ 1,000,001 . . . . add 1
n ≥ log(1,000,001)/log(2) . . . . . take the base-2 logarithm
n ≥ 19.93
The total Angelina receives from Choice 1 will exceed $1,000,000 after 20 days.
First you need to get rid of the parenthesis by distributing the 0.6 to each term inside.
(0.6)(10n) + (0.6)(25) =10+5n
6n +15 = 10+5n subtract the 5n on both sides and subtract 15 from both sides
6n-5n = 10-15
n=-5
Multiplying fractions is simple. You just multiply across so 7/3 x 2/3 is 14/9 and that is the simplest form.
Let number of plastic containers collected by fourth grade= x
Then number of plastic containers collected by fifth grade students=x-216
OR
If number of plastic container collected by fifth grade is y
then number of plastic container collected by fourth grade=y+216
So, we can write it as follows
⇒ number of plastic containers collected by fourth grade= number of plastic containers collected by fifth grade +216
