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xxTIMURxx [149]
3 years ago
15

Compare the graphs of the two functions f and g, given that f(x) = 3x. What is an equation for g? A graph where the x- axis runs

from -6 to 4 and the y- axis runs from -4 to 4. The point (0, 0) where the axes intersect is marked. The points 2, 4, -2, and -4 are marked on the y- axis and the points 2, 4, -2, -4 and -6 are marked on the x- axis. Two graphs of functions f and g are marked in the graph and they overlap each other on the negative x- axis. The function f is mapped in the first and the second quadrants while the function g is mapped in the third and the fourth quadrants. A. g(x)=(−3)x B. g(x)=−3x C. g(x)=3x D. g(x)=(13)−x
Mathematics
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

56

Step-by-step explanation:

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The diameter of the circle would be a line going through the center of the circle connecting to both sides.

The radius of a circle would be a line only going to the the center, or in other words, half the circle.

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3 0
3 years ago
The game of clue involves 6 suspects, 6 weapons, and 9 rooms. one of each is randomly chosen and the object of the game is to gu
mina [271]
Part A:

Given that t<span>he game of clue involves 6 suspects, 6 weapons, and 9 rooms.

The number of ways that one of each is randomly chosen is given by:

^6C_1\times{ ^6C_1}\times{ ^9C_1}=6\times6\times9=324

Therefore, the number of solutions possible is 324.



Part B:

Given that a </span>players is randomly given three of the remaining cards, <span>let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.

The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.

Let x denote the number of solutions that are possible after that player observes his or her three cards, then:

x={ ^{6-s}C_1}\times{ ^{6-w}C_1}\times{ ^{9-r}C_1}=(6-s)(6-w)(9-r)

Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).



Part C:

The expected value E(x) of a data set x_i with probabilities p(x_i) is given by E(x)=\Sigma xp(x)

There are </span>^{3+3-1}C_{3-1}={ ^5C_2}=10 possible combinations s, w and r. They are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)

Thus the expected value is given by

E(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\  \\ = \frac{1}{ ^{21}C_3} (162\cdot{ ^6C_3}\cdot{ ^6C_0}\cdot{ ^9C_0}+162\cdot{ ^6C_0}\cdot{ ^6C_3}\cdot{ ^9C_0}+216\cdot{ ^6C_0}\cdot{ ^6C_0}\cdot{ ^9C_3} \\ \\ +180\cdot{ ^6C_2}\cdot{ ^6C_1}\cdot{ ^9C_0}+192\cdot{ ^6C_0}\cdot{ ^6C_2}\cdot{ ^9C_1}

+210\cdot{ ^6C_1}\cdot{ ^6C_0}\cdot{ ^9C_2}+192\cdot{ ^6C_2}\cdot{ ^6C_0}\cdot{ ^9C_1}+180\cdot{ ^6C_1}\cdot{ ^6C_2}\cdot{ ^9C_0} \\  \\ +210\cdot{ ^6C_0}\cdot{ ^6C_1}\cdot{ ^9C_2}+200\cdot{ ^6C_1}\cdot{ ^6C_1}\cdot{ ^9C_1} \\  \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\  \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\  \\ =\frac{1}{1,330}(264,384) \\  \\ =\bold{198.78}
6 0
3 years ago
TIME REMAINING
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Step-by-step explanation:

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