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Serggg [28]
3 years ago
7

suppose y varies directly with x, an dy = 10 when x =3. what direct variation equations relates x and y?

Mathematics
1 answer:
Ratling [72]3 years ago
8 0

The direct variation equation relates x and y is y = \frac{10}{3} x

Step-by-step explanation:

Direct variation is when one variable is equal to a constant times another variable

If y varies directly with x, then

  • y ∝ x
  • y = k x is the equation of variation
  • k is the constant of variation

∵ y varies directly with x

∵ y ∝ x

∴ y = k x

∵ y = 10 when x = 3

- Substitute these values in the equation above to find k

∴ 10 = k(3)

∴ 10 = 3 k

- Divide both sides by 3

∴ k = \frac{10}{3}

∴ The equation of variation is y = \frac{10}{3} x

The direct variation equation relates x and y is y = \frac{10}{3} x

Learn more:

You can learn more about variation in brainly.com/question/10708697

#LearnwithBrainly

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Solve the following simultaneous equations.<br> 5x – 3y = 6<br> 3x + 4y= 21
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Answer:

y  = 3, x = 3

Step-by-step explanation:

5x – 3y = 6

5x = 6 + 3y

x = (6 + 3y)/5

3x + 4y= 21

3x = 21 - 4y

x = (21-4y)/3

Using simultaneous equations:

(21-4y)/3 = (6 + 3y)/5

--> 105-20y = 18 + 9y

29y = 87

  • y = 3

  • x = (6 + 3(3))/5 = 3
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2 years ago
True or False?
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Answer:

Step-by-step explanation:

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3 years ago
The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)
Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1 .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores  and D = difference between both

So, \mu_A = Population mean for the math scores

       \mu_B = Population mean for the writing scores

 Let \mu_D = Difference between the population mean for the math scores and the population mean for the writing scores.

            <em>  Null Hypothesis, </em>H_0<em> : </em>\mu_A = \mu_B<em>     or   </em>\mu_D<em> = 0 </em>

<em>      Alternate Hypothesis, </em>H_1<em> : </em>\mu_A \neq  \mu_B<em>      or   </em>\mu_D \neq<em> 0</em>

Hence, Test Statistics used here will be;

            \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1    where, Dbar = Bbar - Abar

                                                               s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}

                                                               n = 12

Student        Math scores (A)          Writing scores (B)         D = B - A

     1                      540                            474                                   -66

     2                      432                           380                                    -52  

     3                      528                           463                                    -65

     4                       574                          612                                      38

     5                       448                          420                                    -28

     6                       502                          526                                    24

     7                       480                           430                                     -50

     8                       499                           459                                   -40

     9                       610                            615                                       5

     10                      572                           541                                      -31

     11                       390                           335                                     -55

     12                      593                           613                                       20  

Now Dbar = Bbar - Abar = 489 - 514 = -25

 Bbar = \frac{\sum B_i}{n} = \frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}  = 489

 Abar =  \frac{\sum A_i}{n} = \frac{540+432+528+574+448+502+480+499+610+572+390+593}{12} = 514

 ∑D_i^{2} = 22600     and  s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}} = \sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} } = 37.05

So, Test statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1

                            = \frac{-25 - 0}{\frac{37.05}{\sqrt{12} } } follows t_1_1   = -2.34

<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

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