Which measure of center is best to use to determine the expected number of strokes required for the ninth hole? The mean is the best choice because the data is nearly symmetrical. The median is the best choice because data is clustered. The mean is the best choice because the data is skewed to the right. The median is the best choice because the data is skewed to the left.

The mean is the best choice because the data is nearly symmetrical

Step-by-step explanation:

Question:

The data from the dot plot obtained from another post of the question are;

From the given data, we have that the mean of the 25 players is 4.24

Mean = 4.24 strokes

The median strokes is 4.5 strokes

Median = 4.5

The values of the mean and median are approximately equal because the data are symmetrical and has the normal shape

Therefore, the mean is the best option due to the symmetrical shape of the data.

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

add the numbers under frequency to find the total number:

5 + 8 +7 + 4 + 0 +1 = 25

the answer is 25.