Undefined as the line is parallel to y axis ie; x=3
Answer:
C. 45 and 141 seconds
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 93 seconds
Standard deviation = 16 seconds
99.7% of running times are approximately between:
By the Empirical rule, within 3 standard deviations of the mean, so between 3 standard deviations below the mean and 3 standard deviations above the mean
3 stnadard deviations below the mean
93 - 3*16 = 45 seconds
3 standard deviations above the mean
93 + 3*16 = 141 seconds
The correct answer is:
C. 45 and 141 seconds
Answer:
x₂ = 7.9156
Step-by-step explanation:
Given the function ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)
If f(x) = ln(x)+x-10
f'(x) = 1/x + 1
f(9) = ln9+9-10
f(9) = ln9- 1
f(9) = 2.1972 - 1
f(9) = 1.1972
f'(9) = 1/9 + 1
f'(9) = 10/9
f'(9) = 1.1111
x₁ = x₀ - f(x₀)/f'(x₀)
x₁ = 9 - 1.1972/1.1111
x₁ = 9 - 1.0775
x₁ = 7.9225
x₂ = x₁ - f(x₁)/f'(x₁)
x₂ = 7.9225 - f(7.9225)/f'(7.9225)
f(7.9225) = ln7.9225 + 7.9225 -10
f(7.9225) = 2.0697 + 7.9225 -10
f(7.9225) = 0.0078
f'(7.9225) = 1/7.9225 + 1
f'(7.9225) = 0.1262+1
f'(7.9225) = 1.1262
x₂ = 7.9225 - 0.0078/1.1262
x₂ = 7.9225 - 0.006926
x₂ = 7.9156
<em>Hence the approximate value of x₂ is 7.9156</em>