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3 years ago

How do I solve 4x+8y=20 -4x+2y=-30

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

3 0

Answer:

i got this

Step-by-step explanation:

4x+8y=20 minus 4x both sides but dont minus

-4x -4x

8y=20-4x divide to get y alone

8 8

y= -2x+2.5

know do it to the other side

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soldi70 [24.7K]

Answer: you unsubscribe from it

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3 years ago

The clerk at a store worked 45 hours one week.He worked an equal number of hours each day on Monday through Friday.He worked an

Lubov Fominskaja [6]

Answer:

he worked 40 hours on the the weekdays

Step-by-step explanation:

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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.

3 0

3 years ago

Read 2 more answers

If y varies directly with x and x=4 when, y=20, find k, the constant of variation.

Alborosie

Direct variation means [ y = k x ]

Given . . . 20 = k 4

k = 20/4 = <u>5 .</u>

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3 years ago

Caleb bought a boogie board and used a coupon for $8.50 off of the cost. Fill in the box with the expression that represents the

jenyasd209 [6]

I think this is the answer sorry if I'm wrong
(B)-8.50=T

4 0

3 years ago