The radius of a sphere is claimed to be 3.0 inches, correct to within 0.01 inch. use linear approximation to estimate the resulting error, measured in cubic inches, in the volume of the sphere.

1/4 because there is 4 grades.

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a