A=p(1+i/m)^mn
Interest earned
I=A-p
A=980×(1+0.08÷4)^(4×5)
A=1,456.23
I=1,456.23−980
I=476.23
A=7,200×(1+0.04)^(8)
A=9,853.69
I=9,853.69−7,200
I=2,653.69
A=15,520×(1+0.06÷2)^(2×4)
A=19,660.27
I=19,660.27−15,520
I=4,140.27
Answer:
The limit that 97.5% of the data points will be above is $912.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Find the limit that 97.5% of the data points will be above.
This is the value of X when Z has a pvalue of 1-0.975 = 0.025. So it is X when Z = -1.96.
So
The limit that 97.5% of the data points will be above is $912.
Answer:
x+5/x+3
Step-by-step explanation: