Use compound interest formula F=P(1+i)^n twice, one for each deposit and sum the two results.
For the P=$40,000 deposit,
i=10%/2=5% (semi-annual)
number of periods (6 months), n = 6*2 = 12
Future value (at end of year 6),
F = P(1+i)^n = 40,000(1+0.05)^12 = $71834.253
For the P=20000, deposited at the START of the fourth year, which is the same as the end of the third year.
i=5% (semi-annual
n=2*(6-3), n = 6
Future value (at end of year 6)
F=P(1+i)^n = 20000(1+0.05)^6 = 26801.913
Total amount after 6 years
= 71834.253 + 26801.913
=98636.17 (to the nearest cent.)
Answer:
Step-by-step explanation:
B. Yes, because it passes the vertical line test, is the answer.
If you would like to know how many centimeters are 27 millimeters, you can calculate this using the following steps:
1 cm ... 10 mm
x cm = ? ... 27 mm
1 * 27 = 10 * x
27 = 10 * x /10
x = 27/10
x = 2.7 cm
The correct result would be 2.7 cm.
Answer:
simple even tho im in 6th grade it would be over $100
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:
The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):
Finally, the interval at 98% confidence level is:
