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2 years ago

Identified limitations to the design of a product or system are the

Engineering

1 answer:

kirill [66]2 years ago

5 0

Identified limitations to the design of a product or system are the constraints

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On Beverly's last project, the team identified only a few lessons learned. Which approach to lessons learned

Afina-wow [57]

Option C (making lessons learned a regular part of meetings) is the correct approach.

As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.
As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.

Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

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3 years ago

Please answer me- How do you make friends.? Im a loner. YALL GOTTA HELP ME:(

denpristay [2]

Realize your fear is in your head. The first step is to develop a healthy mental image of meeting new people.

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3 years ago

Read 2 more answers

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the

muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, $\L_e$ = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

$P =\frac{\pi^2EI}{L_e^2}$

where, I is the moment of inertia

on substituting the respective values, we get

$10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}$

I = 13.13 in⁴

also for circular cross-section

I = $\frac{\pi}{64}\times d^4$

thus,

13.13 = $\frac{\pi}{64}\times d^4$

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0

3 years ago

Determine the real roots of f (x) = −0.6x2 + 2.4x + 5.5:(a) Graphically.(b) Using the quadratic formula.(c) Using three iteratio

zubka84 [21]

The three methods used to find the real roots of the function are,

graphically, the quadratic formula, and by iteration.

The correct vales are;

(a) Graphically, the roots obtained are; x ≈ -1.629, and 5.629

(b) Using the quadratic formula, the real roots of the given function are; x ≈ -1.62589, x ≈ 5.62859

Reasons:

The given function is presented as follows;

f(x) = -0.6·x² + 2.4·x + 5.5

(a) The graph of the function is plotted on MS Excel, with increments in the

x-values of 0.01, to obtain the approximation of the x-intercepts which are

the real roots as follows;

$\begin{array}{|c|cc|}x&&f(x)\\-1.63&&-0.00614\\-1.62&&0.03736\\5.62&&0.03736 \\5.63&&-0.00614\end{array}\right]$

Checking for the approximation of x-value of the intercept, we have;

$x = -1.63 + \dfrac{0 - (-0.00614)}{0.0376 - (-0.00614)} \times (-1.62-(-1.63)) \approx -1.629$

Therefore, based on the similarity of the values at the intercepts, the x-

values (real roots of the function) at the x-intercepts (y = 0) are;

x ≈ -1.629, and 5.629

(b) The real roots of the quadratic equation are found using the quadratic

formula as follows;

The quadratic formula for finding the roots of the quadratic equation

presented in the form f(x) = a·x² + b·x + c, is given as follows;

$x = \mathbf{ \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}}$

Comparison to the given function, f(x) = -0.6·x² + 2.4·x + 5.5, gives;

a = -0.6, b = 2.4, and c = 5.5

Therefore, we get;

$x = \dfrac{-2.4\pm \sqrt{2.4^{2}-4\times (-0.6)\times 5.5}}{2\times (-0.6)} = \dfrac{-2.4\pm\sqrt{18.96} }{-1.2} = \dfrac{12 \pm\sqrt{474} }{6}$

Which gives

The real roots are; x ≈ -1.62859, and x ≈ 5.62859

$x_l$ = 5, and $x_u$ = 10

The first iteration is therefore;

$x_r = \dfrac{5 + 10}{2} = 7.5$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{10- 5}{10 + 5} \right | \times 100\% = 33.33\%$

$True \ error, \ \epsilon _t = \left|\dfrac{5.62859 - 7.5}{5.62859} \right | \times 100\% = 33.25\%$

f(5) × f(7.5) = 2.5 × (-10.25) = -25.625

The bracket is therefore; $x_l$ = 5, and $x_u$ = 7.5

Second iteration:

$x_r = \dfrac{5 + 7.5}{2} = 6.25$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{7.5- 5}{7.5+ 5} \right | \times 100\% = 20\%$

$True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 6.25}{5.62859} \right | \times 100\%} \approx 11.04\%$

f(5) × f(6.25) = 2.5 × (-2.9375) = -7.34375

The bracket is therefore; $x_l$ = 5, and $x_u$ = 6.25

Third iteration

$x_r = \dfrac{5 + 6.25}{2} = 5.625$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{5.625- 5}{5.625+ 5} \right | \times 100\% = 5.88\%$

$True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 5.625}{5.62859} \right | \times 100\%} \approx 6.378 \times 10^{-2}\%$

f(5) × f(5.625) = 2.5 × (0.015625) = 0.015625

Therefore, the bracket is $x_l$ = 5.625, and $x_u$ = 6.25

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6 0

3 years ago

Plastic parts produced by an injection-molding operation are checked for conformance to specifications. Each tool contains 11 ca

Wittaler [7]

Answer:

a) 0.27 % b) 0.27%

Explanation:

a) The inspector can choose 1 part from 11 parts in 11C1 ways. These are total number of mutually exclusive events.

Now three cavities are effected by temperature Malfunction that results in the parts that do not conform to specifications. To obtain exactly 1 non conforming part the inspector have to choose from one of the three cavities that are effected by temperature. This can be done in 3C1 ways.

Probability that inspector finds out exactly 1 non conforming part is,

Probability = $\frac{3C1}{11C1}$

Probability = $\frac{\frac{3!}{1!(3-1)!} }{\frac{11!}{1!(11-1)!} }$ = 3 / 11

= 0.27 %

b) As the inspector is choosing only 1 part for inspection so the probability to find exactly 1 non conforming part is same as the probability to find at least 1 non conforming part.

So the probability is 0.27 %.

8 0

4 years ago