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kolezko [41]
3 years ago
6

Suppose the number of gallons of gasoline per day used by a car is normally distributed with a mean of 2.2 gallons and a standar

d deviation of 1.2 gallons.
What is the difference in gallons per day used by a car with a z-score of 3 and another car that has a z-score of 0?

A. 1.2

B. 2.6

C. 3.6

D. 4.6
Mathematics
1 answer:
const2013 [10]3 years ago
5 0
The difference in gallons per day used is 3.6.

The formula to calculate a z-score is:
z=\frac{X-\mu}{\sigma},

where X is the value used to calculate the score, μ is the mean and σ is the standard deviation.  We have the z-scores so we must work backward:

3=\frac{X-2.2}{1.2}\text{ and }0=\frac{X-2.2}{1.2}

For both equations, we will cancel the 1.2 by multiplying both sides:
3\times1.2=(\frac{X-2.2}{1.2})\times1.2\text{ and }0\times1.2=(\frac{X-2.2}{1.2})\times1.2
\\
\\3.6=X-2.2\text{ and }0=X-2.2

Now we will cancel 2.2 from both equations by adding it to both sides:

3.6+2.2=X-2.2+2.2     and       0+2.2=X-2.2+2.2
5.8=X     and       2.2=X

The difference in gas used per day would be given by
5.8-2.2 = 3.6.
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The rod is made of A-36 steel and has a diameter of 0.22 in . If the rod is 4 ft long when the springs are compressed 0.7 in . a
ziro4ka [17]

The force in the rod when the temperature is 150 °F is 718.72 pounds-force.

<h3>How to determine the resulting the resulting force due to mechanical and thermal deformation</h3>

Let suppose that rod experiments a <em>quasi-static</em> deformation and that both springs have a <em>linear</em> behavior, that is, force (F), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to <em>temperature</em> increase creates a <em>spring</em> deformation additional to that associated with <em>mechanical</em> contact.

Given simmetry considerations, we derive an expression for the <em>spring</em> force (F), in pounds-force,  as a sum of mechanical and thermal effects by principle of superposition:

F = k\cdot (\Delta x + 0.5\cdot \Delta l)   (1)

Where:

  • k - Spring constant, in pounds-force per inch.
  • \Delta x - Spring deformation, in inches.
  • \Delta l - Rod elongation, in inches.

The <em>rod</em> elongation is described by the following <em>thermal</em> dilatation formula:

\Delta l = \alpha \cdot L_{o}\cdot (T_{f}-T_{o})   (2)

Where:

  • \alpha - Coefficient of linear expansion, in \frac{1}{^{\circ}F}.
  • L_{o} - Initial length of the rod, in inches.
  • T_{o} - Initial temperature, in degrees Fahrenheit.
  • T_{f} - Final temperature, in degrees Fahrenheit.

If we know k = 1000\,\frac{lb}{in}, \Delta x = 0.7\,in, \alpha = 6.5\times 10^{-6}\,\frac{1}{^{\circ}F}, L_{o} = 48\,in, T_{o} = 30\,^{\circ}F and T_{f} = 150\,^{\circ}F, then the force in the rod at final temperature is:

F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right]

F = 718.72\,lbf

The force in the rod when the temperature is 150 °F is 718.72 pounds-force. \blacksquare

To learn more on deformations, we kindly invite to check this verified question: brainly.com/question/13774755

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3 years ago
A teacher surveys 30 students on the number of books they read over the summer. The data
horrorfan [7]

Answer:

8+12=20÷2=10

median is 10

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