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lubasha [3.4K]
3 years ago
7

I need this math problem done as soon as possible!

Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

See below

Step-by-step explanation:

<u>Parent function:</u>

  • y = 3ˣ

<u>Transformed function:</u>

  • y = 4(3)⁻²ˣ⁺⁸ + 6, (note. I see this as 8, sorry if different but it doesn't make any change to transformation method)

<u>Transformations to be applied:</u>

  • f(x) → f(-x) reflection over y-axis
  • f(-x) → f(-2x) stretch horizontally by a factor of 2
  • f(-2x) → f(-2x + 8) translate 8 units right
  • f(-2x + 8) → 4f(-2x + 8)  stretch vertically by a factor of 4
  • 4f(-2x + 8) → 4f(-2x + 8) + 6 translate 6 units up
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4 : 5 10 : 9

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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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