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Mekhanik [1.2K]
3 years ago
14

Consider the functions f(x) and g(x) shown below. For which of the following values of x does f(x) does not equal g(x)?

Mathematics
1 answer:
olganol [36]3 years ago
6 0
<h3>Answer:  d) -3</h3>

==========================================================

Explanation:

When f(x) = g(x) is true, this is where the two graphs intersect. So we have three solutions here because we have three different intersection points.

The x coordinates of those intersection points, from left to right, are: -3/2, 0, 3.

Note: -3/2 = -1.5 is exactly at the midpoint of -2 and -1.

In other words, if we plugged in x = 0 into both f(x) and g(x), then we'll get the same output value and hence f(x) = g(x) will be true. The same applies to -3/2 and 3 as well.

Recall that y = f(x) and y = g(x), so saying f(x) = g(x) really means they both have the same y outputs for that specific input x value. This explains why intersection points correspond to solutions.

------------

Since x = -3 is not on the list of solutions, this means this x value will make f(x) and g(x) be different outputs. The graph shows the curves do not intersect with each other when x = -3. Instead, g(x) is above f(x) at this location. So we can say g(x) > f(x) when x = -3.

In short, when x = -3, we have f(x) \ne g(x)

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A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th
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Answer:

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the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

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Where;

P(x) is the probability of getting number x on the dice.

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P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

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So, equation 1 becomes;

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The probability of getting either a 5 or a 2 in one throw

P(2U5) = (P(2) + P(5))/P(T)

Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

P(2U5) = 7P(1)/21P(1)

P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

8 0
3 years ago
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