Given Information:
Years = t = 35
Semi-annual deposits = P = $2,000
Compounding semi-annually = n = 2
Interest rate = i = 6.5%
Required Information
Accumulated amount = A = ?
Answer:
Accumulated amount = $515,827
Step-by-step explanation:
The future value of amount earned over period of 35 years and interest rate 6.5% with semi-annual deposits is given by
FV = PMT * ((1 + i/n)^nt - 1)/(i/n))
Where
n = 2
i = 0.065
t = 35
FV = 2000*((1 + 0.065/2)^2*35 - 1)/(0.065/2))
FV = 2,000*(257.91)
FV ≈ $515,827
Therefore, Anthony will have an amount of $515,827 when he retires in 35 years.
Answer: 5 feet.
Step-by-step explanation:
Remember that the lenghts of the sides of a regular octagon are equal.
You can observe in the figure that the lenght of a side of the regular octagon is 1.25 inches.
Let be "s" the actual length of a side of the regular octagon.
Knowing that the scale drawing has a scale of 1 inch 4 feet, you can find the actual lenght of a side of this regular octagon with:

Therefore, the actual length of a side of the octagon is 5 feet.
Answer:
Is D I think
Step-by-step explanation:
Hello there.
<span>Y=6mx-2b what is y equal to
</span>y=<span><span><span>6m</span>x</span>−<span>2<span>b</span></span></span>
Answer:
Population Mean = 2.0
Population Standard deviation = 0.03
Step-by-step explanation:
We are given that the inspector selects simple random samples of 30 finished products and computes the sample mean product weight.
Also, test results over a long period of time show that 5% of the values are over 2.1 pounds and 5% are under 1.9 pounds.
Now, mean of the population is given the average of two extreme boundaries because mean lies exactly in the middle of the distribution.
So, Mean,
=
= 2.0
Therefore, mean for the population of products produced with this process is 2.
Since, we are given that 5% of the values are under 1.9 pounds so we will calculate the z score value corresponding to a probability of 5% i.e.
z = -1.6449 {from z % table}
We know that z formula is given by ;
~ N(0,1)
-1.6449 =
⇒
⇒
0.0608 *
{as sample size is given 30}
⇒
= 0.03 .
Therefore, Standard deviation for the population of products produced with this process is 0.0333.