7.A number of minor automobile accidents occur at various high-risk intersections in Teton County despite traffic lights. The Traffic Department claims that a modification in the type of light will reduce these accidents. The county commissioners have agreed to a proposed experiment. Eight intersections were chosen at random, and the lights at those intersections were modified. At the .01 significance level, is it reasonable to conclude that the modification reduced the number of traffic accidents?

Modification has not reduced the number of accidents.

Step-by-step explanation:

                                 A  B  C   D   E   F  G   H

Before modification 5  7  6   4    8   9   8   10

After modification    3  7  7   0    4   6   8    2

Difference                -2 0  1   -4   -4   -3  0   -8

Mean of differences

Standard deviation of differences =

Formula of paired t test :

Df = n-1 = 8-1 =7

t critical> t calculated

So, We failed to reject null hypothesis

Hence modification has not reduced the number of accidents.

choice b is correct

step-by-step explanation:

we have been given the following system of inequalities;

we are required to graph the solution to the system of inequalities. using a graphing tool, the graph of the system is as shown in the attachment. the continuous line represents the graph of while the dotted line represents .

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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