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ella [17]
3 years ago
5

Can someone help me pls?

Mathematics
1 answer:
Anton [14]3 years ago
3 0
If you put the numbers into a ratio ( 10:15 ) then you would need to solve for ( 20: ? )

Add the terms to produce a whole then put the numbers into fractions using this number as the denominator. 20+10= 30

10:15 20:30

Diameter= 30 mm
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The milligrams of aspirin in a person's body is given by the equation a=500*(3/4) ^t. where t is the number of hours since the p
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Is it c

Step-by-step explanation:

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A student prepares for an exam by studying a list of 10 problems. She can solve 8 of them. For the exam, the instructor selects
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Answer:

Probability to solve 7 problem will be equal to \frac{1}{15}

Step-by-step explanation:

It is given total number of question = 10

She can solve 8 of them from 10 questions

We have to find the probability that she can solve 7 question

Probability will be equal to ratio of favorable outcome to total sample space

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Greatest common factor of 4st,12st
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4 is the greatest common factor.
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4 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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Answer:

B

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The questions says that the graph is shifted down 7 units, which means it will affect the y-intercept.

Remember: If it's shifted up, it's positive & if it's shifted down, it's negative.

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