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1 year ago

Find the solution for this system of equations. 12x 15y = 34 -6x 5y = 3 x = y =

Mathematics

1 answer:

damaskus [11]1 year ago

7 0

The solution for this system of equations. 12x+ 15y = 34 -6x+ 5y = 3

is (x,y) =(2/3, 8/5)

Given a system of equations,

12x + 15y = 34 ------(1)

-6x + 5y = 3 ------(2),

To solve this we use the elimination method

In the elimination method, you either add or subtract the equations to get an equation in one variable.

Equation (1) + 2 × Equation (2),

We get,

(12x+15y=34)+(-12x+10y=6)

⇒ 15y+10y=34+6

⇒ 25y=40

⇒ y=40/25

⇒ y=8/5

From equation (2),

⇒ -6x+5(y)=6

⇒ -6x+5(8/5)=6

⇒ -6x+8=6

⇒ -6x=-2

⇒ x=2/3

Hence the solution for this system of equations. 12x 15y = 34 -6x 5y = 3

is (x,y =(2/3, 8/5)

To learn more about the system of equations visit:

brainly.com/question/27919853

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The probability that a person is sick is: 0.008
The probability that a test is positive, given that the person is sick is 0.9833
The probability that a test is negative, given that the person is not sick is: 0.9899
The probability that a person is sick, given that the test is positive is: 0.4403
The probability that a person is not sick, given that the test is negative is: 0.9998
A 99% accurate test is a correct test

(a) Probability that a person is sick

From the table, we have:

$\mathbf{Sick = 59+1 = 60}$

So, the probability that a person is sick is:

$\mathbf{Pr = \frac{Sick}{Total}}$

This gives

$\mathbf{Pr = \frac{60}{7500}}$

$\mathbf{Pr = 0.008}$

The probability that a person is sick is: 0.008

(b) Probability that a test is positive, given that the person is sick

From the table, we have:

$\mathbf{Positive\ and\ Sick=59}$

So, the probability that a test is positive, given that the person is sick is:

$\mathbf{Pr = \frac{Positive\ and\ Sick}{Sick}}$

This gives

$\mathbf{Pr = \frac{59}{60}}$

$\mathbf{Pr = 0.9833}$

The probability that a test is positive, given that the person is sick is 0.9833

(c) Probability that a test is negative, given that the person is not sick

From the table, we have:

$\mathbf{Negative\ and\ Not\ Sick=7365}$

$\mathbf{Not\ Sick = 75 + 7365 = 7440}$

So, the probability that a test is negative, given that the person is not sick is:

$\mathbf{Pr = \frac{Negative\ and\ Not\ Sick}{Not\ Sick}}$

This gives

$\mathbf{Pr = \frac{7365}{7440}}$

$\mathbf{Pr = 0.9899}$

The probability that a test is negative, given that the person is not sick is: 0.9899

(d) Probability that a person is sick, given that the test is positive

From the table, we have:

$\mathbf{Positive\ and\ Sick=59}$

$\mathbf{Positive=59 + 75 = 134}$

So, the probability that a person is sick, given that the test is positive is:

$\mathbf{Pr = \frac{Positive\ and\ Sick}{Positive}}$

This gives

$\mathbf{Pr = \frac{59}{134}}$

$\mathbf{Pr = 0.4403}$

The probability that a person is sick, given that the test is positive is: 0.4403

(e) Probability that a person is not sick, given that the test is negative

From the table, we have:

$\mathbf{Negative\ and\ Not\ Sick=7365}$

$\mathbf{Negative = 1+ 7365 = 7366}$

So, the probability that a person is not sick, given that the test is negative is:

$\mathbf{Pr = \frac{Negative\ and\ Not\ Sick}{Negative}}$

This gives

$\mathbf{Pr = \frac{7365}{7366}}$

$\mathbf{Pr = 0.9998}$

The probability that a person is not sick, given that the test is negative is: 0.9998

(f) When a test is 99% accurate

The accuracy of test is the measure of its sensitivity, prevalence and specificity.

So, when a test is said to be 99% accurate, it means that the test is correct, and the result is usable; irrespective of whether the result is positive or negative.