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andrezito [222]
1 year ago
11

Please solve this question

Mathematics
1 answer:
ANTONII [103]1 year ago
3 0

The probability that a randomly selected x-value from the distribution will be in the interval:

  • P(35 < x < 45) = 0.6827 and,
  • P(30 < x < 40) = 0.47725

<h3>What is the probability of a normal distribution?</h3>

The probability of a normal distribution can be determined from the symmetrical curve between 1 to 100%.

From the information given:

  • Mean = 40
  • Standard deviation = 5

To determine the probability that a randomly selected x-value is in the given interval:

  • P(35<x<45)
  • P(30<x<40)

P(35 < x < 45) = P(\dfrac{35-40}{5} < Z < \dfrac{45-40}{5})

P(35 < x < 45) = P(\dfrac{-5}{5} < Z < \dfrac{5}{5})

P(35 < x < 45) = P(-1 < Z < 1)

\mathbf{P(35 < x < 45) = P[Z\le 1] -P[Z\le -1]}

Using normal distribution table:

P(35 < x < 45) = 0.8414 - 0.1587

P(35 < x < 45) = 0.6827

P(30 < x < 40) = P(\dfrac{30-40}{5} < Z < \dfrac{40-40}{5})

P(30 < x < 40) = P(\dfrac{-10}{5} < Z < \dfrac{0}{5})

P(30 < x < 40) = P(-2 < Z < 0)

\mathbf{P(30 < x < 40) = P[Z\le0]-P[Z\le -2]}

Using normal distribution table:

P(30 < x < 40) = 0.5 - 0.02275

P(30 < x < 40) = 0.47725

Learn more about the probability of a normal distribution here:

brainly.com/question/4079902

#SPJ1

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To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

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You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

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Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

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for xo = a = 0.235 and yo = f(a) = 0.721:

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The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

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