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masha68 [24]
2 years ago
7

A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution fr

om 10 ounces to 16 ounces, a. what is the average amount of water dispensed by the machine? 13 oz. b. what is the standard deviation of the amount of water dispensed? c. What is the probability that 13 or more ounces will be dispensed in a given glass? d. What is the probability that between 12 and 14 ounces will be dispensed in a given glass?
Mathematics
1 answer:
natima [27]2 years ago
6 0

Answer:

(a) 13 ounces.

(b) 1.732 ounces.

(c) 0.5

(d) 0.33

Step-by-step explanation:

Given : The amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces. So, a=10 and b=16.

(a)  The average amount of water dispensed by the machine = (a+b)/2 = (10+16)/2 = 13 ounces.

(b)  The standard deviation of the amount of water dispensed = \int\limits^a_b {\frac{(b-a)^{2} }{12} } \, dx \\ = √36/12 =√3 = 1.732 ounces.

(c) P (13 or more ounces will be dispensed in a given glass) = \int\limits^a_b {f(x)} \, dx = \int\limits^b_a {\frac{dx}{b-a} } \, dx = 0.5

(d) P (between 12 and 14 ounces will be dispensed in a given glass) = ( here a=12 and b=14.) \int\limits^a_b {f(x)} \, dx = \int\limits^b_a {\frac{dx}{b-a} } \, dx = 0.33

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Step-by-step explanation:

Let's say the number of hours the Carter family used their sprinkler is x and the number of hours the Davis family used their sprinkler for is y.

So combined:

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and we are also told that:

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With some rearranging, we can figure out that:

y= 45 - x

and by substituting that into the second equation:

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Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

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