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scZoUnD [109]
1 year ago
15

The graph of f '(x) is continuous and decreasing with an x-intercept at x = 2. Which of the following statements must be true? (

5 points)
The graph of f is always concave down.
The graph of f is always increasing.
The graph of f has an inflection point at x = 2.
The graph of f has a relative minimum at x = 2.
Mathematics
1 answer:
bixtya [17]1 year ago
5 0

Answer:

The graph of f is always concave down ⇒ the first answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Remember that : If  f(x)  is a function then the solutions to the

 equation f′(x) = 0 gives the maximum and minimum values to f(x)

- The value of  x  gives maximum if f′′(x) is negative and minimum if

  f′′(x) is positive.

- Inflection points of the function  f(x) are found the solutions of the

 equation  f′′(x) = 0

* Lets solve the problem

- The graph of f'(x) is continuous means that the graph is unbroken line

- The graph of f'(x) decreasing with an x-intercept at x = 2 means

 f'(2) = 0

- The differentiation of a function equal to zero at the critical point

  (minimum or maximum) of the function

∵ f'(x) = 0 at x = 2

∴ The x-coordinate of the critical point of f(x) is 2

- If the differentiation of the function is decreasing, then the critical

 point of the function is maximum point

∵ f'(x) is decreasing

∴ The critical point of the f(x) is maximum point

- That means the slope of curve is negative

∴ The graph of f is concave down at x = 2

* The right answer is the graph of f is always concave down

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Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = .0116

Step-by-step explanation:

a.

    E(X) = np = .40 probability * 136 trials = 54.4 blocked transmissions

    To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

    i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

   ii. E(X) = 1/p , according to the geometric model.

              = 1/.4 = 2.5.

       We expect that the specialist will review 2.5 suspicious transactions <em>on average </em>before finding the first transmission that will be blocked.

c.

    i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

    ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

        P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

                    = 45 (.4^2)(.6^10-2) = .0016

        As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

        Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

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Answer:

h=7.65

Step-by-step explanation:

H is directly proportional to the square root of p;

Let k be the constant of proportionality;

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This means for corresponding points of h and p such that (h1,p1) and (h2,p2) we have;

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