Let Y be a random variable with a mean of 11 and variance 9. Using Chebyshev’s theorem to find the value of C such that P(|Y −11| ≥ C) ≤ 0.09.

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1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

the answer is y = 11.54

hope this : )

10 b/c 5*2 =10

step-by-step explanation: