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3 years ago

14

Can someone help me with this I’ll mark as brainliest.

2 answers:

nikdorinn [45]3 years ago

7 0

1 = 10
2 = 12
3 = 14
4 = 16
5 = 18
6 = 20

Δweight
------------- =
Δ cost

2-1
------- =
12-10

1
—
2

vodka [1.7K]3 years ago

3 0

Answer:

The W/C is divided by the numbers on top to the number on the bottom

may i have brainlist??

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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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3 years ago

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