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never [62]
3 years ago
14

If the cost of a competing factor of production, such as a machine that also could do the job, rises. This would result in a shi

ft in the demand for this factor out to the right and would put pressure on the wage to rise. I don't understand why this is the case because I thought if machines replaced human labor then wouldn't wages be lower because there wouldn't be a need for human labor?
Mathematics
1 answer:
Harrizon [31]3 years ago
6 0

Keep in mind that we're framing it based on what the first sentence says, which is "If the cost of a competing factor of production, such as a machine that also could do the job, rises".

So if the cost of getting a machine part, various parts, or the entire machine cost rises, then demand for the machine will go down. This will make employers seek out substitutes. In this case, those substitutes would be human labor. As employers demand for labor goes up, the wages will rise assuming the supply of workers is held constant. If the supply of workers increased, then you could argue the wages could go down. So that's why I'm assuming the supply is held in check.

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An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

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⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0
2 years ago
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WARRIOR [948]

Answer:

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Answer:

Solution given:

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opening bracket

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