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asambeis [7]
2 years ago
10

State whether or not the relation is considered a function?

Mathematics
1 answer:
WINSTONCH [101]2 years ago
3 0

Answer:

A function is when each input has exactly one output

Step-by-step explanation:

Each input can only have one output, but an output can have multiple inputs.

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Answer:

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Step-by-step explanation:

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A firm is hiring labor and capital in the cost-minimizing combination. Which of the following would cause the firm to increase h
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The productivity of labor increases by 5% would cause the firm to increase hiring capital and decrease hiring of labor

<u>Step-by-step explanation:</u>

A company can convert input data to output data. So, it gets divided the data into broad categories called production factors. Productivity variations, or more precisely the marginal products of labour, work in the same way as differences in real wages. Remember that marginal prices depend on both - real wages and the productivity.

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4 years ago
Convert 400 percent into a decimal
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3 years ago
The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris a
nadezda [96]

Answer and explanation:

Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

and the probabilities of poor print quality given printers are

P(E|A)=0,\ P(E|B)=0.3,\ P(E|C)=0.4,\ P(E|D)=0.6

First we calculate the probability that print quality is poor,

P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)

P(E)=(0)(0.8)+(0.3)(0.02)+(0.4)(0.08)+(0.6)(0.1)

P(E)=0+0.006+0.032+0.06

P(E)=0.098

a. Determine the probability of high ink viscosity given poor print quality.

P(C|E)=\frac{P(E|C)P(C)}{P(E)}

P(C|E)=\frac{0.4\times 0.08}{0.098}

P(C|E)=\frac{0.032}{0.098}

P(C|E)=0.3265

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

P(A|E)=\frac{P(E|A)P(A)}{P(E)}

P(A|E)=\frac{0\times 0.8}{0.098}

P(A|E)=\frac{0}{0.098}

P(A|E)=0

Probability of misaligned paper given poor quality is

P(B|E)=\frac{P(E|B)P(B)}{P(E)}

P(B|E)=\frac{0.3\times 0.02}{0.098}

P(B|E)=\frac{0.006}{0.098}

P(B|E)=0.0612

Probability of printer-head debris given poor quality is

P(D|E)=\frac{P(E|D)P(D)}{P(E)}

P(D|E)=\frac{0.6\times 0.1}{0.098}

P(D|E)=\frac{0.06}{0.098}

P(D|E)=0.6122

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

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