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zalisa [80]
2 years ago
11

The coefficient of determination is used to: compute correlations for truncated ranges. compare the relative strength of coeffic

ients. identify curvilinear relationships. test correlations for causality.
Mathematics
1 answer:
kozerog [31]2 years ago
6 0

Answer: compare the relative strength of coefficients.

Step-by-step explanation: The Coefficient of determination usually denoted as R^2 is obtained by taking the squared value of the correlation Coefficient (R). It's value ranges from 0 to 1 and the value obtained gives the proportion of variation in the dependent variable which could be attributed to it's correlation or relationship to th independent variable. With a R^2 value close to 1, this means a large portion of Variation in a variable A could be explained due to changes in variable B while a low value signifies a low variance between the variables. Hence, the Coefficient of determination is used in comparing the relative strength of the Coefficients in other to establish whether a weak or strong relationship exist.

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Triphasil-28 birth control tablets are taken sequentially, 1 tablet per day for 28 days, with the tablets containing the followi
bazaltina [42]

Answer:

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

Step-by-step explanation:

The total milligrams of levonorgestrel is:

L = L_{1} + L_{2} + L_{3}

In which L_{1}, L_{2} and L_{3} are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

E = E_{1} + E_{2} + E_{3}

In which E_{1},E_{2} and E_{3} are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

6*0.050mg = 0.30mg of levonorgestrel and 6*0.030mg = 0.18mg of ethinyl estradiol are taken.

So L_{1} = 0.30 and E_{1} = 0.18

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

5*0.075mg = 0.375mg of levonorgestrel and 5*0.040mg = 0.20mg of ethinyl estradiol are taken.

So L_{2} = 0.375 and E_{2} = 0.20

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

10*0.125mg = 1.25mg of levonorgestrel and 10*0.030mg = 0.30mg of ethinyl estradiol are taken.

So L_{3} = 1.25 and E_{3} = 0.30

The total milligrams of levonorgestrel is:

L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925

The total milligrams of ethinyl estradiol is:

E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

6 0
2 years ago
Identifying functions from relations
Mashutka [201]
I don’t know but I need the Points sorry for you
4 0
1 year ago
Select all of the following that are ordered pairs of the given function.
Tatiana [17]
F(x) is the same as y.......so basically ur subbing in ur points into the equation to see if it comes out equal.

f(x) = 3 - 2x.....(-2,-1)....x = -2 and f(x) = -1
-1 = 3 - 2(-1)
-1 = 3 + 2
-1 = 5.....this is not true, so it is not a solution

and that is how to do this problem.....

(-1,5)......this IS a solution
(0,3)......this IS a solution
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8 0
3 years ago
A cylinder has a volume of 390 cm³. If a cone has the same dimensions as the cylinder then what is the volume of the cone.
dangina [55]

Answer: The volume of the cone is 130 cubic centimeters.

Step-by-step explanation:

We know that to find the volume of a cylinder you will use the formula

V= \pir^{2} *h  Where r is the radius and h is the height.

And to find the volume of a cone you will use the formula

V= \pir^{2}*h /3

So if they have the same dimensions which is the height and the radius then we will just divide the  volume of the cylinder by 3 to find the volume of the cone.

V = 390/3

V =  130 cm^3

8 0
3 years ago
Find the integral of 3/sqrt of 1-4x^2
erastovalidia [21]
\int {\frac{3}{\sqrt{1-4x^{2}}}} \, dx
= 3\int {\frac{1}{\sqrt{1-4x^{2}}}} \, dx
= 3\int {\frac{1}{\sqrt{4(\frac{1}{4} - x^{2})}}} \, dx
= \frac{3}{2}\int {\frac{1}{\sqrt{\frac{1}{4} - x^{2}}}}\, dx

= \frac{3}{2}sin^{-1}(2x) + C
8 0
3 years ago
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