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Lunna [17]
3 years ago
7

Pls help. i never understood this rule.

Mathematics
1 answer:
Dimas [21]3 years ago
8 0
L'Hopital rule only applies to indeterminate cases of 0/0 or ∞/∞ or 0*0 or ∞ * ∞.

and you'd need to make the expression a rational, where the numerator and denominator gives you 0/0 or ∞/∞, then you apply L'Hopital.

the only candidate there is 

\bf \lim\limits_{x\to \infty}\cfrac{2x^2-1}{3x+1}\implies \lim\limits_{x\to \infty}\cfrac{2(\infty)^2-1}{3(\infty)+1}\implies \cfrac{\infty}{\infty}
\\\\\\
\textit{so we can use L'Hopital by instead using the derivatives}
\\\\\\
\lim\limits_{x\to \infty}\cfrac{2x^2-1}{3x+1}\stackrel{LH}{\implies }\lim\limits_{x\to \infty}\cfrac{4x}{3}\implies \cfrac{4(\infty)}{3}\implies \infty
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X - 5 = 20 <br> x= <br><br> I need to know what the x equals
Monica [59]

Answer:

x = 25

Step-by-step explanation:

Given

x - 5 = 20 ( isolate x by adding 5 to both sides )

x = 25

5 0
3 years ago
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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
Nuetrik [128]

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

5 0
3 years ago
Which of the following gives a single solution of the |x+3|- 2
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Answer:

x=-1

Step-by-step explanation:

given \left | x+3 \right |-2=0

\left | x+3 \right |=2

x+3=+2

x=-1,     also,  x+3=-2

                    x=-2-3=-5

hence x=-1 , -5 answer

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Answer:

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Step-by-step explanation:

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Not enough information is that all of the question?
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