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SIZIF [17.4K]
1 year ago
10

The graph below shows two polynomial functions, f(x) and g(x): 6 f(x) 5 V -3 -2 2 3 4 g(x) -4 -5 -7 Which of the following state

ments is true about the graph above? (4 points) Of(x) is an even degree polynomial with a negative leading coefficient. g(x) is an even degree polynomial with a negative leading coefficient. Of(x) is an odd degree polynomial with a positive leading coefficient. O g(x) is an odd degree polynomial with a positive leading coefficient.

Mathematics
1 answer:
olya-2409 [2.1K]1 year ago
7 0

The function f(x) is a parabola. We can tell it is a even degree polynomial because the axis of reflection is a vertical line (it passes through the vertex).

In this case, f(x) has a double root at (0,1), but a parabola will have up to 2 roots.

It has a positive leading (quadratic in this case) coefficient, because it is concave up.

In the case of g(x), we can tell it is an odd degree polinomial, as it has a axis of reflection that is a line with slope m=-1. It is like it is reflected two times, one on an horizontal line and then on a vertical line.

The leading coefficient is positive because g(x) tends to infinity when x increases, and the leading coefficient is the one that has more weight for large values of x, so it has to be positive to have positive values of g(x).

Then, we can go through the statements.

O f(x) is an even degree polynomial with a negative leading coefficient. FALSE (the leading coefficient is positive)

O g(x) is an even degree polynomial with a negative leading coefficient. FALSE (it is an odd degree polynomial)

O f(x) is an odd degree polynomial with a positive leading coefficient. FALSE (it is an even degree polynomial).

O g(x) is an odd degree polynomial with a positive leading coefficient.​ TRUE

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b. < D = 64°

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The National Highway Traffic Safety Administration reported the percentage of traffic accidents occurring each day of the week.
Zigmanuir [339]

Answer:

Hence,

a)

Test statistic X^{2} = 9.269.  

p value = 0.148

b)

Sunday=  15.23%, Monday= 12.62%, Tuesday= 13.10%, Wednesday=11.67%,

Thursday= 13.10%, Friday= 16.19%, Saturday= 18.09%.

Step-by-step explanation:

a)

                    relative observed    Expected       Chi square

Category   frequency(p)      O_{i}    Ei=total*p    (O-E)^{2} R^{2}_{i} =(O_{i} -E_{i} )^{2} /E_{i}

     1           0.142857143      64.0         60.00    16.00          0.267

    2           0.142857143      53.0       60.00     49.00         0.817

    3           0.142857143      55.0         60.00     25.00        0.417

    4           0.142857143      49.0       60.00     121.00        2.017

    5           0.142857143      55.0         60.00     25.00        0.417

    6           0.142857143      68.0         60.00     64.00        1.067

    7           0.142857143      76.0         60.00    256.00        4.267

  Total         1.000      420        420             556       9.269

Test statistic X^{2} = 9.269.  

p value = 0.148 from excel: chidist(9.269,6)  

b)

Percentage %= Frequency of traffic occur on category x 100 / Total number of frequency given.

Category          Percentage%

Sunday                   15.23%

Monday                    12.62%

Tuesday                   13.10%

Wednesday           11.67%

Thursday           13.10%

Friday                   16.19%

Saturday                   18.09%

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