Hole? It depends how big it is. Since there is no hole being depicted in this question, the answer is inexcusably 2ft
Answer:
a) $20,771.76
b) $20,817.67
c) $20,484.80
d) $20,864.52
Step-by-step explanation:
<u>Compound Interest Formula</u>

where:
- A = final amount
- P = principal amount
- r = interest rate (in decimal form)
- n = number of times interest applied per time period
- t = number of time periods elapsed
<u>Part (a): semiannually</u>
Given:
- P = $15,000
- r = 5.5% = 0.055
- n = 2
- t = 6 years
Substitute the given values into the formula and solve for A:



<u>Part (b): quarterly</u>
Given:
- P = $15,000
- r = 5.5% = 0.055
- n = 4
- t = 6 years
Substitute the given values into the formula and solve for A:



<u>Part (c): monthly</u>
Given:
- P = $15,000
- r = 5.5% = 0.055
- n = 12
- t = 6 years
Substitute the given values into the formula and solve for A:



<u>Continuous Compounding Formula</u>

where:
- A = Final amount
- P = Principal amount
- e = Euler's number (constant)
- r = annual interest rate (in decimal form)
- t = time (in years)
<u>Part (d): continuous</u>
Given:
- P = $15,000
- r = 5.5% = 0.055
- t = 6 years
Substitute the given values into the formula and solve for A:


Learn more about compound interest here:
brainly.com/question/27747709
brainly.com/question/28004698
Answer:
0.3125
Step-by-step explanation:
If it decreases by 50% each time it hits the ground then all you have to do is divide 10 by 2 (50%) each time it bounces.
10 / 2 = 5 (first bounce)
5 / 2 = 2.5 (second bounce)
2.5 / 2 = 1.25 (third bounce)
Ect. . .
Until on the fifth bounce it is 0.3125m high.
Answer:
No, the conditions for normality have not been met because the sample size for the quarters is not large enough and no information is given about the distributions of the populations.
Step-by-step explanation:
"The Kolmogorov-Smirnov test (K-S) and Shapiro-Wilk (S-W) test are designed to test normality by comparing your data to a normal distribution with the same mean and standard deviation of your sample. If the test is NOT significant, then the data are normal, so any value above .05 indicates normality"
Answer:
8:6 & 12:9 & 16:12
Step-by-step explanation:
hope this help ☆