Classify the relationships and justify your choice in each case. Assume the independent variable is mentioned first
a) A strong positive correlation exists between students’ marks in Kitchener high
schools and the city’s air quality index.
b) There is a positive correlation between the earth’s average air temperature and
the concentration of CO2 (carbon dioxide) in the atmosphere
c) Human alcohol consumption and level of coordination have a strong negative
correlation
d) Interest in televised sport is positively correlated with fitness level
e) Movie box office sales are negatively correlated with ticket price
f) Highway traffic congestion is positively correlated with the number of
urban expressways.
types of relationships:
Cause and effect relationship is the correlation between two variables in which
a change in one directly causes a change in the other
Common cause relationship is the correlation between two variables in which both
variables change as a result of a third common variable
Presumed relationship exists when two variables are related but no causality can
be inferred.
Reverse cause and effect relationship occurs when the assumed causation
becomes reversed.
Accidental relationship is a relationship that is based purely on coincidence.

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

the number of the people going.

step-by-step explanation:

the cost to buy one movie ticket is $7. if the total cost for the movie is a function of how many people go, then the input is : the number of the people going.

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:

i don't see any

step-by-step explanation: