How to find the constant of variation for the quadratic variation?

1 answer:

7 0

Put the values you know in the formula and solve for the constant of variation.
.. y = kx^2
.. k = y/x^2 . . . . . fill in values for y and x and compute.

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Dawn comma Sergio comma Tyrone comma and Jim have all been invited to a dinner party. They arrive randomly and each person arriv

professor190 [17]

Answer:

a. 24

b. 2

c. 0.0833 = 8.33%

Step-by-step explanation:

The first "slot" of person to arrive has 4 possibilities, then the second "slot" will have 3 possibilities, as one has already arrived, then the third "slot" has 2 possibilities, and the fourth "slot" has just 1 possibility.

So, multiplying all these combinations, we have 4*3*2*1 = 24 possible ways they can arrive

If the first and the last person are already "locked", we just have possibilities for the second and third person. The second will have 2 possibilities (Sergio or Tyrone), and the third will have only 1 (the person that wasn't the second between Sergio and Tyrone). So, the number of possibilities is 2*1 = 2

If we have 2 cases where Dawn is first and Jim is last, from a total of 24 possible cases, the probability is 2/24 = 1/12 = 0.0833 = 8.33%

3 0

3 years ago

Use the following table for questions 11-13.A baseball manager believes a linear relationship exists between the number of Home

slamgirl [31]

The regression equation of Y on X is given by the following formula:

$Y-\bar{Y}=b_{yx}(X-\bar{X})$

Where byx is given by the formula:

$b_{yx}=\frac{N\sum^{}_{}XY-\sum^{}_{}X\sum^{}_{}Y}{N\sum^{}_{}X^2-(\sum^{}_{}X)^2}$

Where N is the number of values (N=8). We need to find the sum of X values, the sum of Y values, the average of X, the average of Y, the sum of X*Y and the sum of X^2.

The table of values is:

The values we need to know are on the following table:

By replacing the known values in the formula we obtain:

$\begin{gathered} b_{yx}=\frac{8\cdot26125-167\cdot995}{8\cdot4649-(167)^2} \\ b_{yx}=\frac{209000-166165}{37192-27889} \\ b_{yx}=\frac{42835}{9303} \\ b_{yx}=4.6 \end{gathered}$

Now, the average of X and Y is the sum divided by N, then: