Bianca has a choice between three savings accounts. Savings account A has an APR of 3.9998%, compounded semiannually; savings ac
count B has an APR of 3.9742%, compounded monthly; and savings account C has an APR of 3.9886%, compounded quarterly. Bianca wants to choose the savings account with the highest APY. (5 points: Part I – 1 point; Part II – 1 point; Part III – 1 point; Part IV – 1 point; Part V – 1 point) Part I: If r is a savings account's APR and n is the number of compounding periods per year, what expression gives the savings account's APY?
1 answer:
Answer:
APY = (1 +r/n)^n -1
Step-by-step explanation:
Each period the principal is multiplied by 1 plus the periodic rate: 1 + r/n. For n periods in a year, the effective multiplier is (1 +r/n)^n. The value after this multiplication includes the principal, so to find the interest, we need to subtract the principal. The effective rate is ...
APY = (1 +r/n)^n -1
_____
You didn't ask, but the effective rates for the accounts shown are given in the attachment. The formula bar shows the formula used, and the highlighted cell shows the highest APY.
(I find a spreadsheet is a simple way to do repetitive calculations with minimal chance of error.)
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Answer:
Option B
Step-by-step explanation:
The coordinates are (-1,8) and (3,-4)
Slope =
Slope =
Slope =
Slope = -3
Answer:
We accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.
Step-by-step explanation:
We have these following hypothesis:
Null
Equal means
So
Alternative
Different means
So
Our test statistic is:
In which are the sample means,
are the sample sizes and
are the standard deviations of the sample.
In this problem, we have that:
So
What to do with the null hypothesis?
We will reject the null hypothesis, that is, that the means are equal, with a significante level of if
In which v is the number of degrees of freedom, given by
Applying the formula in this problem, we have that:
So, applying t at the t-table at a level of 0.975, with 20 degrees of freedom, we find that
We have that
Which is lesser than t.
So we accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.