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galben [10]
3 years ago
10

Which functions are linear? ax+by=c y=kx y=x^2

Mathematics
1 answer:
9966 [12]3 years ago
8 0

Answer: Ax+By=C, y=kx

Step-by-step explanation:

Ax+By=C is the standard form of a line.

y=kx is a proportional line, which is just a linear equation without the intercept.

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Which of the following triangles have three sides of different length? A. acute B. scalene C.equilateral D. right​
katen-ka-za [31]

Answer:

scalene

hope this helps

have a good day :)

Step-by-step explanation:

4 0
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.

8 0
3 years ago
Find the approximate value of the circumference of a circle with the given radius. use = 3.14. round your results to one more de
sveta [45]
The circumference of the circle is about 25.12 inches. Hope it help!
4 0
3 years ago
Solve for n.<br><br> n + 1 = 4(n – 8)
kolezko [41]
N + 1 = 4(n – 8)
Use distributive property for this part: 4(n – 8) ----> 4n - 32

n + 1 = 4n - 32

Get the variable on one side: 

n + 1 = 4n - 32
-n         -n

1= 3n - 32

Get the variable on its own:

1= 3n - 32 
+32       +32

33 = 3n

Divide by 3 on both sides: 

33/3 = 3n/3

n = 11



7 0
3 years ago
Solve the equation.
djverab [1.8K]
(y - 5)/3 = 1
y - 5 = 3 x 1 = 3
y = 3 + 5 = 8

6x + 29 = 5
6x = 5 - 29 = -24
x = -24/6 = -4
7 0
3 years ago
Read 2 more answers
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