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. a. Determine the probability of high ink viscosity given poor print quality.
b. Given poor print quality, what problem is most likely?

Answers

Answer 1

Answer:

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.

$P(C|E)=(P(E|C)P(C))/(P(E))$

$P(C|E)=(0.4* 0.08)/(0.098)$

$P(C|E)=(0.032)/(0.098)$

$P(C|E)=0.3265$

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

Probability of no printer problem given poor quality is

$P(A|E)=(P(E|A)P(A))/(P(E))$

$P(A|E)=(0* 0.8)/(0.098)$

$P(A|E)=(0)/(0.098)$

$P(A|E)=0$

Probability of misaligned paper given poor quality is

$P(B|E)=(P(E|B)P(B))/(P(E))$

$P(B|E)=(0.3* 0.02)/(0.098)$

$P(B|E)=(0.006)/(0.098)$

$P(B|E)=0.0612$

Probability of printer-head debris given poor quality is

$P(D|E)=(P(E|D)P(D))/(P(E))$

$P(D|E)=(0.6* 0.1)/(0.098)$

$P(D|E)=(0.06)/(0.098)$

$P(D|E)=0.6122$

From the above conditional probabilities,

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

Answer 2

Answer:

Answer:

Answer of Part(a) is 16/49

and Answer of Part(b) is Printer-head debris

Step-by-step explanation:

Answer is in the following attachment

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On a test a student got 15 problems correct out of a total of 20 problems.What percent did the student get incorrect?
HURRY

Answers

Divide 15 by 20

15 divided by 20 is 75%

Final answer:

To calculate the percentage of incorrect answers, subtract the correct answers from the total amount, resulting in 5 incorrect answers. Then divide the incorrect answers by the total problems, multiplied by 100, giving us 25% incorrect.

Explanation:

The subject is the calculation of percentages, specifically determining what percent of the test problems a student got incorrect. To find the answer, first subtract the number of problems the student got correct (15) from the total number of problems on the test (20). This gives you 5 incorrect answers. Then, to convert this number into a percentage, divide the number of incorrect problems (5) by the total number of problems (20) and multiply by 100. This gives an answer of 25%. So, the student got 25% incorrect on the test.

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A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is no net force on the system, what is the velocity of the pitcher? −0.1 m/s −0.2 m/s −0.7 m/s −1.4 m/s

Answers

Answer:

The velocity of the pitcher is −0.1 m/s

Step-by-step explanation:

Given : Mass of pitcher = 50 kg

Mass of Baseball= 0.15kg

Velocity of Ball = 35m/s

To Find : velocity of the pitcher

Solution :

The total momentum of the system is conserved when no external force acts on a system .The total initial momentum of the system is equal to the total final momentum of the system.

Since , the ball and the pitcher are initially at rest, therefore, the total initial momentum of the system is zero.

Since we are given that no external forces act on the system , the total final momentum of the system is also equal to zero.

Let us suppose the mass of the pitcher is $m_(p)$

Speed of pitcher = $v_(p)$

The mass of the ball is $m_(b)$

Speed of ball = $v_(b)$

So, the final momentum of the system of pitcher and the ball is given by:

$momentum =m_(p) v_(p) +m_(b) v_(b) =0$

⇒ $50* v_(p) +0.15*35 =0$

⇒ $50* v_(p) +5.25 =0$

⇒ $v_(p) = (-5.25)/(50)$

⇒ $v_(p) = -0.105$

Thus , The velocity of the pitcher is -0.105m/s≈−0.1 m/s

Negative sign shows the opposite direction.

Hence The velocity of the pitcher is −0.1 m/s

the answer is -0.1 m/s if you're looking for the Edgnuity answer. I just took the test. Hope this helps!

Select the correct answer.What is the current value of a future sum of money called?
А.
current value
B.
present value
С. .
future value

Answers

Answer:

b. present money

Step-by-step explanation:

the concept that States an amount of money today is worth more than that sum amount in the future. future money is not worth much then the amount received today.

Answer is B, Present Value

An ordinary regression model that treats the response Y as normally distributed is a special case of a GLM, with normal random component and identity link function. 1. With a GLM. Y does not need to have a normal distribution and one can model a function of the mean of Y instead of just the mean itself, but In order to get the maximum likelihood estimates the variance of Y must be constant at all values of predictors. ii. The Pearson residual e_inty_i-muhatiysqrt(muhat_) for a GM has a large-sample standard normal distribution (a)) True False, (w) True; (b)) True 00 True True (co False 0 false, Oll) False; IdFalse 00 True (1) False;

Answers

An ordinary Regression model that treats the response Y is (a) True False, (w) True

What is Regression?

A statistical method called regression links a dependent variable to one or more independent (explanatory) variables.

A regression model can demonstrate whether changes in one or more of the explanatory variables are related to changes in the dependent variable.

A) Models for numerical response variable, like ANOVA and linear regression are special cases of GLMs

for these model the following holds

1. Random component has a normal distribution

2. Systematic component α+β₁x₁+β₂x₂+...........βₓxₓ

3. link function = identity (g(µ)=µ)

GLMs can generalise these models with response Y as normally distributed, hence the statement is True

B) With a GLM. Y does not need to have a normal distribution and one can model a function of the mean of Y instead of just the mean itself. but in order to get ML estimates the variance of Y must be small. This small variance of Y is the reason for ML estimator to be the best one. hence the statement is false.

An ordinary Regression model that treats the response Y is (a) True False, (w) True

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Suppose that 2121% of all births in a certain region take place by Caesarian section each year. a. In a random sample of 500500 births, how many, on average, will take place by Caesarian section? b. What is the standard deviation of the number of Caesarian section births in a sample of 500500 births? c. Use your answers to parts a and b to form an interval that is likely to contain the number of Caesarian section births in a sample of 500500 births.

Answers

Answer:

a) $E(X) = np=500*0.21= 105$

b) $Sd(X) = √(82.95)= 9.108$

c) Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

$\mu -2\sigma = 105-2*9.108=86.785$

$\mu +2\sigma = 105+2*9.108=123.215$

So we expect about 86 and 123 most of the numbers of Caesarian section births

Step-by-step explanation:

For this case we can define the random variable X as the number of births in the Caesarian section and from the data given we know that the distribution of X is:

$X \sim Binom (n = 500, p=0.21$

Part a

The expected value for this distribution is given by:

$E(X) = np=500*0.21= 105$

Part b

The variance is given by:

$Var(X) = np(1-p) = 500*0.21*(1-0.21)= 82.95$

And the deviation would be:

$Sd(X) = √(82.95)= 9.108$

Part c

Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

$\mu -2\sigma = 105-2*9.108=86.785$

$\mu +2\sigma = 105+2*9.108=123.215$

So we expect about 86 and 123 most of the numbers of Caesarian section births

1/4 times 2 equals??

Answers

Answer:

1/2

Step-by-step explanation:

Answer:

1/2

Step-by-step explanation:

1/4 = 0.25 x 2 = 0.5 = 1/2

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