Answer: the value of the account at the end of 6 years is is $8577
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = 6000
r = 6% = 6/100 = 0.06
n = 4 because it was compounded 4 times in a year.
t = 6 years
Therefore,.
A = 6000(1+0.06/4)^4 × 6
A = 6000(1+0.015)^24
A = 6000(1.015)^24
A = $8577
Answer:
Here we will use algebra to find three consecutive integers whose sum is 300. We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 300. Therefore, you can write the equation as follows:
(X) + (X + 1) + (X + 2) = 300
To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:
X + X + 1 + X + 2 = 300
3X + 3 = 300
3X + 3 - 3 = 300 - 3
3X = 297
3X/3 = 297/3
X = 99
Which means that the first number is 99, the second number is 99 + 1 and the third number is 99 + 2. Therefore, three consecutive integers that add up to 300 are 99, 100, and 101.
99 + 100 + 101 = 300
We know our answer is correct because 99 + 100 + 101 equals 300 as displayed above.
Step-by-step explanation:
Answer:
One
General Formulas and Concepts:
<u>Algebra I</u>
- Reading a coordinate plane
- Solving systems of equations by graphing
Step-by-step explanation:
If 2 lines are parallel, they will have no solution.
If 2 lines are the same, they will have infinite amount of solutions.
We see from the graph that the 2 lines intersect at one point, near x = 1.5.
∴ our systems has 1 solution.
Correction:
Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on
P(E∩E') = P(E)P(E').
Answer:
The case is not always true.
Step-by-step explanation:
Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.
And for any two mutually exclusive events, E and E',
P(E∩E') = 0
Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then
P(E)P(E') cannot be equal to zero.
So
P(E)P(E') ≠ 0
This makes P(E∩E') different from P(E)P(E')
Therefore,
P(E∩E') ≠ P(E)P(E') in this case.
