Does this measure of central tendency describe the data well? You work for a pet-grooming service. You record the number of animals you groomed each day over a 10-day period: 7, 10, 8, 9, 7, 11, 8, 6, 7, and 9. You state that the typical number of animals that you can groom is the mean of 8.2 animals per day.

O Yes, because the mean always describes data well.

O Yes, because there are no outliers to skew the data.

O No, because it is not possible to groom 0.2 of an animal.

O No, because the most common number of dogs groomed is 7.

O Yes, because half of the time, it is possible to groom more than 8.2 and half the time less than 8.2.

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.

55 because 55 + 55 = 110 good luck babe

answer: 1.06 km per sec

step-by-step explanation:

39.6 m=63.72 km

60 min in an hour so 63.72/60=1.06

a) at most means it can cost $12 or less

so the inequality sign would be ≤

you would then need to multiply the number of toppings by the 0.35 for each one and add it to the cost of the cheese pizza.

the equation would be 9.99 + 0.35x ≤ 12

b) using the above equation:

9.99 + 0.35x ≤ 12

subtract 9.99 from each side:

0.35x ≤2.01

divide both sides by 0.35:

x ≤ 2.01 / 0.35

x ≤5.74

at most you could have 5 toppings.

the 5 topping pizza would cost: 0.35 x 5 = 1.75 + 9.99 = $11.74