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.
__
<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.
__
<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.
Whisker plots is not that hard you just have to no the Quartiles
the center of the graph is Q2 the first dot to the left is the lower quartile the last dot to the left is upper quartile so the second dot is Q1 and the 4 dot is Q3
Answer:
55
Step-by-step explanation:
Remark
This is a really good question to know the answer to. <PRQ = 1/2 POQ (the central angle. )
The central angle for this question is 110o. So any angle that has its vertex on the circumference is 1/2 110 = 55. The central angle and the angle on the circumference must be related as they are as this question is. (Both are on the same side of PQ which is not drawn but you can draw it).
Answer:
The answer would be C. 5
Step-by-step explanation:
