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Flura [38]
1 year ago
5

A lateral thoracic spine on a 33-cm patient is usually taken using 100 mA (large focal spot), 0.5 seconds, 86 kVp, 40-inch SID,

12:1 grid ratio. If the mA is increased to 400 to stop motion blur from tremors, which of the following technique changes will produce a radiographic density and contrast most similar to the original?​
Mathematics
1 answer:
Pie1 year ago
4 0

A 0.13 sec and a 86 kVp technical changes are the changes that would have to produce the radiographic density and the contrast that are more similar to the original.

<h3>What is the radiographic density?</h3>

The radio graphic density can de defined to be the total amount or the overall darkening that can be found in a particular radiograph. This is usually known to have a certain type of density range that lies between  0.3 to 2.0 density.

When there is a density that is less than 0.3, the problem is basically because there is the issue of the density which would be found at the base and the fact that the film that is being used has fog in it.

It is known that based on the values that we have here the density and contrast would have to be of the given kvp of 86 and 0.13 seconds in order to be said to be similar to the original.

Read more on the radiographic density here:

brainly.com/question/14332593

#SPJ1

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inessss [21]
The answer is x = 85/36.
7 0
3 years ago
The market and Stock J have the following probability distributions:
denis-greek [22]

Answer:

1) E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%

2) E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%

3) E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1

And the variance would be given by:

Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89

And the deviation would be:

Sd(M) = \sqrt{13.89}= 3.73

4) E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8

And the variance would be given by:

Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56

And the deviation would be:

Sd(M) = \sqrt{55.56}= 7.45

Step-by-step explanation:

For this case we have the following distributions given:

Probability  M   J

0.3           14%  22%

0.4           10%    4%

0.3           19%    12%

Part 1

The expected value is given by this formula:

E(X)=\sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%

Part 2

E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%

Part 3

We can calculate the second moment first with the following formula:

E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1

And the variance would be given by:

Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89

And the deviation would be:

Sd(M) = \sqrt{13.89}= 3.73

Part 4

We can calculate the second moment first with the following formula:

E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8

And the variance would be given by:

Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56

And the deviation would be:

Sd(M) = \sqrt{55.56}= 7.45

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3 years ago
Determine whether the relation is a function. find the domain and range
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What’s the question
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3 years ago
Will mark brainly :)
WARRIOR [948]

Answer:

If is positive, then the parabola opens upward, so the function decreases on and increases on . But if is negative, then just the reverse

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2 years ago
Determine if
Gnesinka [82]

The values of a and b are 1/4 and 1/5 respectively

<h3>How to determine the true statement?</h3>

The function is given as:

16x^2 - 25y^2 = 1

The above function is a horizontal hyperbola.

A horizontal hyperbola that passes through the origin is represented as:

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

Rewrite 16x^2 - 25y^2 = 1 as

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By comparison, we have:

a^2 = \frac{1}{16}

b^2 = \frac{1}{25}

Solve for  a and b

a = \frac{1}{4}

b = \frac{1}{5}

Hence, the values of a and b are 1/4 and 1/5 respectively

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brainly.com/question/27799190

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