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Eddi Din [679]
4 years ago
13

On a game show, Elena is given the choice of three doors: Behind one door is the Grand Prize; behind the others, consolation pri

zes. She picks Door A, and the host, who knows what is behind each door, opens Door B, revealing a consolation prize. The host then offers Elena the opportunity to change her selection to Door C. What should she do?
A. She should not accept because the probability of Door A being the Grand Prize increased.
b. She should accept because the probability of Door C being the Grand Prize increased.
c. She could either accept or not accept because now the probabilities of both doors are equal.
d. She could either accept or not accept because the probabilities of the three doors were equal from the beginning.
Mathematics
2 answers:
user100 [1]4 years ago
7 0

Answer:

c

Step-by-step explanation:

vivado [14]4 years ago
6 0

Answer:

c. She could either accept or not accept because now the probabilities of both doors are equal.

Step-by-step explanation:

After door B was opened, the probability that the prize is behind door A is 50% and behind C is also 50%, so, it doesn't matter which door she chooses.

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Vera_Pavlovna [14]

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rotation of 270 counterclock wise

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Find the quotient for 7.868+1.4
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3 years ago
F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

6 0
3 years ago
Acellus
aksik [14]

9514 1404 393

Answer:

  {-8, -4, -2, -1, 1, 2, 4, 8}

Step-by-step explanation:

Possible rational roots are positive and negative divisors of the constant 8:

  {-8, -4, -2, -1, 1, 2, 4, 8}

_____

<em>Additional comment</em>

This polynomial has no rational roots. Its two real roots are both negative and irrational.

7 0
3 years ago
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