First question:
interest rate is 5% annually. After 4 years relative value of interest will be 1.05^4
Now we multiply that with starting funds and get
4000* 1.05^4 = 4862.03
Second question:
Because interest is 8% semiannually that means that after 35 years relative value of interest will be 1.04^(35*2)
6000*1.04^70 = 93,429.71
Third question:
after first year tractor's value will be 0.86*15450. After 3 years its value will be:
(0.86)^3*15,450 = 9,827.07
1) 4 times 4 = 16
2) 2 times 2 times 9 = 36
3) 4 times 2 times 9 = 72
4) x + y + z = n
(any solution that is true)
we have
Min=26
Q1=67
Q2=80
Q3=87
Max=100
Mean=76
Mode=100
standard deviation=76
IQR=20
so
In this problem
the mean is equal to the standard deviation
therefore
<h2>The standard deviation is incorrectly</h2>
3 times 2 is 6 times 5 is 35
Because the vertex of the parabola is at (16,0), its equation is of the formy = a(x-10)² + 15
The graph goes through (0,0), thereforea(0 - 10)² + 15 = 0100a = -15a = -0.15
The equation is y = f(x) = -0.15(x - 10)² + 15
The graph is shown below.
Part A
Note that y = f(x).
The x-intercepts identify values where the function or y=0. The x-intercepts occur at x=0 and x=20, or at (0,0) and (20,0).
The maximum value of y occurs at the vertex (10, 15) because the curve is down due to the negative leading coefficient of -0.15.
The curve increases in the interval x = (-∞, 10) and it decreases in the interval x = (10, ∞).
Part B
When x=12, y = -0.15(12 - 10)² + 15 = 14.4When x=15, y = -0.15(15 - 10)² + 15 = 11.25
The average rate of change between x =12 to x = 15 is(11.25 - 14.4)/(15 - 12) = -1.05
This rate of change represents the slope of the secant line from A to B. It approximates the rate at which f(x) decreases in the interval x =[12, 15].
