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ollegr [7]
3 years ago
11

How do I show that this function is one-to-one algebraically?

B3%7D%20%2B8" id="TexFormula1" title="f(x)= (x-2)^{3} +8" alt="f(x)= (x-2)^{3} +8" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Gwar [14]3 years ago
6 0
For a one-to-one function, f(x) = f(y).
 So, lets find f(x) and f(y).

We know, 
f(x) = (x - 2)³ + 8

Now,
f(y) = (y - 2)³ + 8

Now, we said earlier, for a function to be one-to-one, f(x) = f(y).

Therefore,

            f(x) = f(y)
(x - 2)³ + 8 = (y - 2)³ + 8
      (x - 2)³ = (y - 2)³     
          x - 2 = y - 2
               x = y

Since we got x = y, we know for every x, there is one and only one y. Therefore, the function is one-to-one function.
 
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-3 1/2 - 2 1/3

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-1/2 - 1/3? u need to find the LEAST common denominator

-3/6 - 2/6 = -5/6

how did I get 6?

Rewriting input as fractions if necessary:
1/2, 1/3

For the denominators (2, 3) the least common multiple (LCM) is 6.

Therefore, the least common denominator (LCD) is 6.

Now lastly combine them total and its -5 - 5/6 = -5 5/6.


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1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n
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Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 70% of fatalities involve an intoxicated driver, hence p = 0.7.
  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

Hence

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061

P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186

P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700

P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916

P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305

P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

Then:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

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