9 coins that equal 76 cents
2 answers:
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Answer:
The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.
P(6.8 ≤X≤9.3) = 0.5932
Step-by-step explanation:
Step 1:-
by using normal distribution find the areas of given x₁ and x₂
Given The average watermelon weighs 8 lbs
μ = 8
standard deviation σ = 1.5
I) when x₁ = 6.8lbs and μ = 8 and σ = 1.5
ii) when x₂ = 9.3 lbs and μ = 8 and σ = 1.5
<u>Step2</u>:-
The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.
P(6.8 ≤X≤9.3) = A(z₂) - A(-z₁)
= A(0.866) - A(-0.8)
= A(0.866)+ A(0.8)
check below normal table
= 0.3051 + 0.2881
= 0.5932
<u>Conclusion</u>:-
The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.
P(6.8 ≤X≤9.3) = 0.5932
Answer:
a)
x = -1/√3 = -0.58 is a local maximum
x = 1/√3 = 0.58 is a local minimum
b)
(-1/√3 , 1/√3 ) DECREASING
(-∞, -1/√3) U (1/√3, ∞) INCREASING
Step-by-step explanation:
To answer all of the questions we must obtain the derivative of the fucntion:
If
U(x) = 4(x^3 - x)
then
U'(x) = 4(3x^2 - 1)
U''(x) = 4(3*2 x) = 24 x
U'''(x) = 24
The local maxima and minima of the function U(x) can be found when U'(x) = 0
this occurs when :
3x^2 = 1
that is:
x = ±1/√3
We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as <u><em>Second derivative test</em></u>), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.
Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.
a)
x = -1/√3 = -0.58 is a local maximum
x = 1/√3 = 0.58 is a local minimum
b)
the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }
The first derivative will be positive when:
3x^2 > 1
|x| > 1/√3 --> x > 1/√3 and x < -1/√3
The first derivatice will be negative when:
3x^2 < 1
|x| < 1/√3 --> x < 1/√3 and x > -1/√3
Therefore the intervals are:
(-1/√3 , 1/√3 ) DECREASING
(-∞, -1/√3) U (1/√3, ∞) INCREASING
<u><em>** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum</em></u>
