Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism, which he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. Employee

Before

After

1

7
7

2

7
1

3

2
6

4

5
4

5

7
3

6

6
3

7

4
1

8

1

4

At the .05 significance level, can he conclude that the number of absences has declined? Estimate the p-value.
a. State the decision rule for 0.1 significance level: H0 : μd ≤ 0; H1 : μd > 0. (Round the final answer to 3 decimal places.)
Reject H0 if t > .
b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)
Value of the test statistic
c. Estimate the p-value? (Round the final answer to 4 decimal places.)
d. What is your decision regarding H0?
At the 0.05 significance level, (Click to select)do not rejectreject H0.

2500

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: