Lfador produces gear for a racquetball league. it costs him s0.005 per square centimeter to racquetballs for the league. in the past, the league used racquetballs with diameter 4 cm, and he produced 30,000 racquetballs. however, league regulations now require racquetballs to have diameter 4.05 cm. use differentials to estimate the increase in cost that alfador must suffer in order to produce 30,000 racquetballs under its new regulations. should this be an overestimate or an underestimate? you may use the fact that the surface area of a ball satisfies sx2, were x is the diameter of the ball

out estimation is underestimate when we increase the cost.

Step-by-step explanation:

Given diameter of the racquetballs (x)= 4 cm

New diameter = 4.05 cm

Therefore, change in diameter ( Δ x) = 0.05.

Cost Function of x diameter of 30,000 racquetballs at the rate 0.005 per square centimeter

C(x) = 30000 × 0.005 × S/4

C(x) = 150 × S/4 = 37.5 × S.

We are given S = π r^2

Therefore,

C(x)= 37.5 π r^2.

C(4) = 37.5 *π*(4)^2 = 37.5 π*16

C (4+0.05) = 37.5 *π*(4.05)^2 = 37.5 * π * 16.4025.

Finding first derivative of cost function, we get

C'(x) = 2 * 37.5 * π r = 75 π r

C(4) = 75 * π * 4 = 300 π

We know,

Δ C(x)/ Δ x = C'(x).

Plugging value of Δ x and x

Δ C(4) = Δ x * C '(4) = 0.05 * 300 π = 15 π = 15 * 3.141 = 47.12.

We also know

Δ C(x) = C(x+Δx) - C(x) = (37.5 * π * 16.4025) - (37.5 π*16) = 37.5 π (16.4025 -16)

                            = 37.5 π(0.4025)

                            = 37.5 * 3.14 * 0.4025

                            = 47.42.

We can see that 47.42 is greater than 47.12.

Therefore, out estimation is underestimate when we increase the cost.

two data sets with the same median will not have the same varaibility

step-by-step explanation:

50-60=10

step-by-step explanation: