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aleksandr82 [10.1K]
3 years ago
14

Which equation is represented by the table

Mathematics
2 answers:
faltersainse [42]3 years ago
7 0
The 1st equation ( y = 1 - 2x) represents the table:
PROOF:
f(1) =1-2 = -1
f(2) = 1-4 = - 3
f(3) = 1-6 = - 5 , etc..
----
----
Crank3 years ago
6 0
One way of solving this is simply by substituting for each x in the table into each formula to see if you get the corresponding y in the table

so if x is 1, 
then in the first equation -> y = 1-2(1) = y= 1-2 = -1
so right off the bat, we found the right equation
but essentially you should do it until you find the one that fits
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Step-by-step explanation:

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Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

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\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

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                   \sf = \dfrac{-\pi }{3}

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For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

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Answer:

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Step-by-step explanation:

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Of the choices listed, the one attached is the rational function.

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SOLUTION:

Step 1:

From the given question, and comparing the scores in classes A and B

Step 2:

Question one, to know the class which had the better overall result on the exam;

(I) The scores in class A ranges from 65 to 100, while that of class B ranges from 60 to 90. This explains that class A had better results.

(ii) The median of the scores in class A is 85, while the median of the scores in class B is 75, this is another piece of supporting evidence that class A had better results.

Step 3:

Question two, to know the class which had greater variability in the results;

For a better understanding of this question, I need to explain the concept of variability in statistics.

Variability in statistics refers to the difference being exhibited by data points within a data set, as related to each other or as related to the mean. This can be expressed through the range, variance or standard deviation of a data set.

Step 4:

So we need to find the range of scores in each of the two classes and then compare, the class with the greater range has the greater variability.

The Range is the difference between the lowest and highest score (H - L), where H is the highest score and L is the lowest score.

Step 5:

Applying the formula for range stated in step 4;

For Class A; H = 100 and L = 65

For Class B; H = 90 and L = 60

The range for class A; H - L = 100 - 65 = 35

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By comparing the range of class A and that of B, it is clear that Class A had a greater range (variability)

CONCLUSION:

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