Nina made two investments: Investment \text{A}A has a value of \$50$50 at the end of the first year and increases by 8\%8% per y
ear. Investment \text{B}B has a value of \$60$60 at the end of the first year and increases by \$3$3 per year. Nina checks the value of her investments once a year, at the end of the year. What is the first year in which Nina sees that investment \text{A}A's value exceeded investment text{B}B's value?
3 answers:
Answer:
year 7
Step-by-step explanation:
If we assume that investment A earns interest compounded annually, its value can be modeled by the equation ...
A = 50·(1+0.08)^(t-1) . . . . . where t is the year number
The second investment earns $3 per year, so its value can be modeled by the equation ...
B = 60 + 3(t -1) . . . . . . . . . where t is the year number
We are interested in finding the minimum value of t such that ...
A > B
50·1.08^(t-1) > 60 +3(t-1)
This is a mix of exponential and polynomial terms for which no solution method is available using the tools of Algebra. A graphing calculator shows the solution to be ...
t > 6.552
The value at the end of year 1 is found for t=1, so the values of interest are seen after 6.55 years, in year 7.
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Answer:
test statistic (Z) is 2.5767 and p-value of the test is .009975
Step-by-step explanation:
: percentage of students who smoke did not change
: percentage of students who smoke has changed
z-statistic for the sample proportion can be calculated as follows:
z= where
- p(s) is the sample proportion of smoking students (
=0.25)
- p is the proportion of smoking students in the survey conducted five years ago (18% or 0.18)
- N is the sample size (200)
Then, z= ≈ 2.5767
What is being surveyed is if the percentage of students who smoke has changed over the last five years, therefore we need to seek two tailed p-value, which is .009975.
This p value is significant at 99% confidence level. Since .009975 <α/2=0.005, there is significant evidence that the percentage of students who smoke has changed over the last five years