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Flura [38]
3 years ago
9

A student analyzed the table below and stated the following, “the y-intercept is 4 because (4,0) has a y value of 0”. Is the stu

dent correct? Why or why not?

Mathematics
1 answer:
Butoxors [25]3 years ago
5 0

Answer:

No

Why he is not correct is because the y-intercept is the coordinate of the point where the graph meets or crosses the y-axis, which is equivalent to a point on the graph where the x-coordinate value reduces to zero, or to put it in a mathematical form x = 0

Step-by-step explanation:

From the given data, we have;

x     {}       y

2     {}      -2

4     {}       0

6      {}      2

8      {}      6

The given data is linear because it has a constant first (common) difference of 2

The general form of the straight line equation is y = m·x + c

Where;

m = The slope

c = The y-intercept

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

The slope of the graph of the data from any two points, (x₁, y₁) = (4, 0) and (x₂, y₂) = (4, 0)) is m = (2 - 0)/(6 - 4) = 1

The equation representing the data in point and slope form is therefore;

y - 4 = 1 × (x - 0)

Which gives;

y = x + 4

Therefore, the y-intercept = 4.

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Step-by-step explanation:

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2. According the exponents propertiess, when you have a multiplication of two powers that has the equal base, you must add the exponents. THerefore, you have the following result:

 (\frac{2}{5}a^{2}x)(-\frac{1}{5}x^{2}a)=-\frac{2}{25}a^{3}x^{3}

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the length of a rectangle field is 7 m less than 4 times the width. The perimeter is 136 m. Find the width and length of the rec
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Let width be x m 

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A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood. Does this suggest that the ac
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Answer:

We  reject H₀   Porcentage of type A blood  donations differs from 40% (porcentage of population having type A blood)

Step-by-step explanation:

We are going to develop a proportion test.

Information we have:

Proportion   P₀   =  40 %      P₀   =  0,4  (porcentage of population having type A blood)

Sample size    n   =   150

Sample  mean   P  =  82/ 150      P  = 0,5466

As  P₀  =   0,4      Q₀  =  0,6     P₀*Q₀   =  0,24

1.-Test Hypothesis:

H₀    null hypothesis                            P₀   =   0.4

Hₐ   alternative hypothesis                 P₀   ≠   0.4

2.- signficance level

     a  )   α  =  0.01    we have a two tail-test      α/2    =   0.005  

     b  )   α  =  0.05                                               α/2    =   0.025

Then  from t-student table we get t(c)   n  = 150    df =  149

     a  )   α/2    =   0.005             t(c)    =  2.581

     b  )   α/2    =   0.025             t(c)    =  1.962

3.-Compute t(s)

t(s)   =   (  P  -  P₀ )  / √P₀Q₀/n  

Plugging  in known values  

t(s)   =[ ( 0.5466 -  0,4  )*√150 ] / √0.24

t(s)   = 0,1466 * 12.25 / 0.4899

t(s)   =  3.6657

Compare t(s)  and  t(c)

t(s)  >  t(c)       3.6657  >  2.581      

Then  t(s) is out of the acceptance region  we reject  H₀.

Simple inspection led us see that for

α/2    =   0.025             t(c)    =  1.962

The situation is the same

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