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Olin [163]
3 years ago
8

Which factors influence changes in consumer demands? check all that apply

Engineering
2 answers:
hoa [83]3 years ago
8 0

Hello!

The factors that affect consumer demand for a product is the price and the person's income.

If the price of the product increases, the demand for the product will decrease. If the price of the product decreases, then the demand for the product will increase.

Demand is also affected by a person's income. The more money a person makes, the more money he has to spend on the product, which increases demand. If a person makes less money than he did before, demand for a product will decrease because he can no longer afford it.

I hope this helps you! Have a lovely day!

- Mal

Simora [160]3 years ago
8 0

Answer:

Here is the right answer: Elasticity, Clearance sales and income.

Hope it helps.

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A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
ILL GIVE BRAINLIEST!!!
Sedbober [7]
Get the app socratic I saw the answer to your question on the app but I ran out of screen time to show you
6 0
3 years ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
Radda [10]

Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

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4 years ago
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Yanka [14]

Answer:

what kind of content? I need to know

4 0
3 years ago
A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of
schepotkina [342]

Answer:

n = 2.36

Explanation:

The stress experimented by the circular bar is:

\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)

\sigma = 10.186\,kpsi

The safety factor is:

n = \frac{24\,kpsi}{10.186\,kpsi}

n = 2.36

5 0
3 years ago
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