Anna’s bank gives her a loan with a stated interest rate of 10.22%. How much greater will Anna’s effective interest rate be if t
2 answers:
Answer:
d. 0.0463% points.
Step-by-step explanation:
We have been given that Anna's bank gives her a loan with a stated interest rate of 10.22%.
To find the how mush the interest rate is effective, while compounded daily than compounded monthly, we will find the difference between interest compounded daily from interest compounded monthly.
Let us convert our given rate in decimal form.
Upon substituting our given interest rate in above equation we will get,
Let us convert our rate in percentage by multiplying our answer by 100.
Therefore, the Anna's effective interest rate will be 0.04633% points and option d is the correct choice.
greater will Anna’s effective interest rate be if the interest is compounded daily, rather than compounded monthly. Option (d) is correct.
Further explanation:
Given:
The options are as follows,
(a). 0.5389 percentage points.
(b). 0.1373 percentage points.
(c). 0.4926 percentage points.
(d). 0.0463 percentage points.
Explanation:
The relationship between the effective interest rate for one year and for effective interest rate for n times in a year can be calculated as follows,
The difference between the effective interest rate be if the interest is compounded daily, rather than compounded monthly can be calculated as follows,
greater will Anna’s effective interest rate be if the interest is compounded daily, rather than compounded monthly. Option (d) is correct.
Option (a) is not correct.
Option (b) is not correct.
Option (c) is not correct.
Option (d) is correct.
Learn more:
- Learn more about inverse of the function brainly.com/question/1632445.
- Learn more about equation of circle brainly.com/question/1506955.
- Learn more about range and domain of the function brainly.com/question/3412497
Answer details:
Grade: High School
Subject: Mathematics
Chapter: Simple interest
Keywords: nominal rate, compounded daily, effective rate, compounded monthly, Anna’s bank, Anna, bank, loan, percentage, interest rate, Principal, invested, interest rate, account, effective interest rate, total interest, amount.
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The forcast for a period
is given by the formular
where
is the actual value for the preceding period and 
is the forcast for the preceding period.
Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.80
Part 1B:
</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.78</span>
Part 2A:
Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.70
</span>
<span><span>Part 2B:
</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.68
</span></span>
<span>Part 3A:
Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the forecast for period 11 is $1.65
</span>
<span><span>Part 3B:
</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the forecast for period 12 is $1.64
Part 4:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157
</span><span><span>Part 5:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107
</span></span>
<span><span>Part 6:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>
The forcast for a period
where
Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.80
Part 1B:
</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.78</span>
Part 2A:
Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.70
</span>
<span><span>Part 2B:
</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.68
</span></span>
<span>Part 3A:
Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the forecast for period 11 is $1.65
</span>
<span><span>Part 3B:
</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the forecast for period 12 is $1.64
Part 4:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157
</span><span><span>Part 5:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107
</span></span>
<span><span>Part 6:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>