A cylindrical vase has a diameter of 6 inches. At the bottom of the vase, there are 8 marbles, each of diameter 3 inches. The vase is filled with water up to a height of 10 inches. Which of the following could be used to calculate the volume of water in the vase? (6 points) π(10in)2(3in) − 1.5( four over three π(8in)3) π(3in)2(10in) − 1.5( four over three π(8in)3) π(10in)2(3in) − 8( four over three π(1.5in)3) π(3in)2(10in) − 8( four over three π(1.5in)3) I got the answer as the fourth one however im unsure if I am correct.

  π(3 in)²(10 in) -8(4/3)(1.5 in)³

Step-by-step explanation:

The volume of a cylinder with a radius of 3 inches and a height of 10 inches is ...

  V = πr²h = π(3 in)²(10 in)

The volume of 8 spheres with a radius of 1.5 inches is ...

  V = 8(4/3)πr³ = 8(4/3)π(1.5 in)³

The volume occupied by the water is the volume of the cylinder not occupied by the spheres. So, it is ...

  V = π(3 in)²(10 in) -8(4/3)(1.5 in)³ . . . . . . matches the 4th choice

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.