Answer:
The value remain unchanged.
Step-by-step explanation:
By Z score, we mean:
Z = 
, if we substitute the given values of mean and standard deviation, we obtain the z score value. That is,
Z = 
= 85/12 = 7.083333.
NB: We assume that mean and standard deviation are in the same units.
In linear models there is a constant additve rate of change. For example, in the equation y = mx + b, m is the constanta additivie rate of change.
In exponential models there is a constant multiplicative rate of change.
The function of the graph seems of the exponential type, so we can expect a constant multiplicative exponential rate.
We can test that using several pair of points.
The multiplicative rate of change is calcualted in this way:
[f(a) / f(b) ] / (a - b)
Use the points given in the graph: (2, 12.5) , (1, 5) , (0, 2) , (-1, 0.8)
[12.5 / 5] / (2 - 1) = 2.5
[5 / 2] / (1 - 0) = 2.5
[2 / 0.8] / (0 - (-1) ) = 2.5
Then, do doubt, the answer is 2.5
<span>Given the Keynesian equation C=A+MD, we find 10000 + (0.8 x 60000) = $58000. M is the Marginal propensity to consume, the A is the Autonomous consumption, and D is disposable income, giving Annual consumer spending as C.</span>
Answer:
On occasions you will come across two or more unknown quantities, and two or more equations
relating them. These are called simultaneous equations and when asked to solve them you
must find values of the unknowns which satisfy all the given equations at the same time.
Step-by-step explanation:
1. The solution of a pair of simultaneous equations
The solution of the pair of simultaneous equations
3x + 2y = 36, and 5x + 4y = 64
is x = 8 and y = 6. This is easily verified by substituting these values into the left-hand sides
to obtain the values on the right. So x = 8, y = 6 satisfy the simultaneous equations.
2. Solving a pair of simultaneous equations
There are many ways of solving simultaneous equations. Perhaps the simplest way is elimination. This is a process which involves removing or eliminating one of the unknowns to leave a
single equation which involves the other unknown. The method is best illustrated by example.
Example
Solve the simultaneous equations 3x + 2y = 36 (1)
5x + 4y = 64 (2) .
Solution
Notice that if we multiply both sides of the first equation by 2 we obtain an equivalent equation
6x + 4y = 72 (3)
Now, if equation (2) is subtracted from equation (3) the terms involving y will be eliminated:
6x + 4y = 72 − (3)
5x + 4y = 64 (2)
x + 0y = 8
