Present Value of an annuity is given by the formular
PV = P(1 - (1 + r)^-n)/r; where PV = $28,000, r = 0.081/12 = 0.00675, n = 35 and P is the periodic (monthly) payment.
P = PVr/(1 - (1 + r)^-n) = (28,000 x 0.00675)/(1 - (1 + 0.00675)^-35) = 189/0.2098 = 900.90
Therefore, the monthly payment is $900.90
1. 4x + 2y = 11
x - 2 = -2y
First I would isolate one of the variables (x or y) of one of the equations, and then substitute it into the other equation.
The easiest to isolate is the "x" in the second equation
x - 2 = -2y Add 2 on both sides
x = -2y + 2
Substitute this into the first equation
4x + 2y = 11
4(-2y + 2) + 2y = 11 Multiply 4 into (-2y + 2)
-8y + 8 + 2y = 11 Combine like terms
-6y + 8 = 11 Subtract 8 on both sides
-6y = 3 Divide -6 on both sides
y = -3/6 Simplify
y = -1/2
Now that you know "y", you can plug it into either of the original equations to find "x"
x - 2 = -2y
x - 2 = -2(-1/2)
x - 2 = 1 Add 2 on both sides
x = 3
Answer is A
2. y = 3x + 5
4x - y = 5
Substitute the first equation into the second equation
4x - y = 5
4x - (3x + 5) = 5 Multiply/distribute the - into (3x + 5)
4x - 3x - 5 = 5 Combine like terms
x - 5 = 5 Add 5 on both sides
x = 10
Plug in "x" into either of the original equations to find "y"
y = 3x + 5
y = 3(10) + 5
y = 30 + 5
y = 35
Answer is A
The GCF of 120 and 36 is 12
Answer:
C. Mrs. Alvarez's scores were less spread out than Mr. Crawford's scores.
Step-by-step explanation:
Mean Absolute Deviation is one of the Statistical measures which we can you to determine the variation that exist amongst a given set of data
Mean Absolute Deviation can be defined as how far or the distance between one set of data to another set of data.
The smaller the Mean Standard Deviation, the lower the degree of variation in the set of data. The data is less spread out
The larger the Mean Standard Deviation, the higher the degree of variation in the set of data. The data is Largely spread out
We are told in the question that:
Mrs. Alvarez's scores had a lower mean absolute deviation than Mr. Crawford's scores. Our conclusion would be that Mrs. Alvarez's scores were less spread out than Mr. Crawford's scores.
Option 2 is correct.
