Here is the region of integration of the integral integral from negative 6 to 6 integral from x squared to 36 integral from 0 to 36 minus y dz dy dx. rewrite the integral as an equivalent integral in the following orders. a. dy dz dx by. dy dx dz c. dx dy dz d. dx dz dy e. dz dx dy

a) ∫_{-6}^{6} ∫_{0}^{36} ∫_{x²}^{36} (-y) dy dz dx

b)  ∫_{0}^{36} ∫_{-6}^{6} ∫_{x²}^{36} (-y) dy dx dz

c)  ∫_{0}^{36} ∫_{x²}^{36} ∫_{-6}^{6} (-y) dx dy dz

e) ∫_{x²}^{36} ∫_{-6}^{6} ∫_{0}^{36} (-y) dz dx dy

Step-by-step explanation:

We write the equivalent integrals for given integral,

we get:

a) ∫_{-6}^{6} ∫_{0}^{36} ∫_{x²}^{36} (-y) dy dz dx

b)  ∫_{0}^{36} ∫_{-6}^{6} ∫_{x²}^{36} (-y) dy dx dz

c)  ∫_{0}^{36} ∫_{x²}^{36} ∫_{-6}^{6} (-y) dx dy dz

e) ∫_{x²}^{36} ∫_{-6}^{6} ∫_{0}^{36} (-y) dz dx dy

We changed places of integration, and changed boundaries for certain integrals.

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5