How do you do this question?

9.29

Step-by-step explanation:

S₄ is the area using Simpson's rule and 4 intervals.

Simpson's rule can be calculated as:

Sᵢ = Δx/3 (f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ... + 4f(xᵢ₋₂) + 2f(xᵢ₋₁) + f(xᵢ))

Or, Simpson's rule can be calculated as a combination of midpoint and trapezoid sums:

S₂ᵢ = (2Mᵢ + Tᵢ) / 3

Using the first method:

Δx = (4−0)/4 = 1

S₄ = 1/3 (1.00 + 4(1.41) + 2(2.24) + 4(3.16) + 4.12)

S₄ = 1/3 (27.88)

S₄ ≈ 9.29

Using the second method:

S₄ = (2M₂ + T₂) / 3

The midpoint area for 2 intervals is:

M₂ = (2) (1.41) + (2) (3.16) = 9.14

The trapezoid area for 2 intervals is:

T₂ = ½ (1.00 + 2.24) (2) + ½ (2.24 + 4.12) (2) = 9.60

Therefore:

S₄ = (2 (9.14) + 9.60) / 3

S₄ ≈ 9.29

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]