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Eddi Din [679]
2 years ago
15

Solve the system by substitution

Mathematics
1 answer:
kumpel [21]2 years ago
6 0
<h3>Answer:</h3>

  (x, y) = (1, 4)

<h3>Step-by-step explanation:</h3>

<u>Given:</u>

  • -4.5x-2y=-12.5
  • 3.25x-y=-0.75

<u>Find:</u>

  x and y using the method of substitution

<u>Solution:</u>

To solve a system of equations by <em>substitution</em>, you find an expression for one of the variables in terms of the others, then substitute that wherever that variable is used*. Usually, you choose <em>one of the equations</em> to solve for the variable you're going to substitute for, then you make the substitution into the <em>remaining</em> equation(s).

Since the coefficient of y in the second equation is -1, it is convenient to solve for y in that equation, then use the resulting expression to substitute for y in the first equation.

  3.25x = y - 0.75 . . . . . . add y

  3.25x + 0.75 = y . . . . . . add 0.75

Using this expression in place of y in the first equation, we have ...

  -4.5x -2(3.25x +0.75) = -12.5

  -11x -1.5 = -12.5 . . . . . . . simplify

  -11x = -11 . . . . . . . . . . . . . add 1.5

  x = 1 . . . . . . . . . . . . . . . . . divide by -11

We can substitute this value into the equation we have for y:

  y = 3.25x +0.75 = 3.25·1 + 0.75

  y = 4

The solution to the system of equations is (x, y) = (1, 4).

_____

* There's no point in substituting for the variable in the equation you used to find the expression. It will give you no useful information. Here, that would look like ...

  3.5x -(3.5x+0.75) = 0.75 . . . . . . . . substitute for y in the second equation

  0.75 = 0.75 . . . . . . . always true. Not a useful substitution.

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Answer:

                           Drug User   Not a drug user    Totals

Test positive       475                     475                  950

Test negative      25                      9025               9050

Totals                   500                   9500               10,000

Step-by-step explanation:

                           Drug User   Not a drug user    Totals

Test positive       475                     475                  950

Test negative      25                      9025               9050

Totals                   500                   9500               10,000

As we know that 5% of all people are drug users

Total drug users=5% of 10,000=0.05(10,000)=500

Total non drug users= Total people - drug users= 10,000-500=9500

or total non drug users=95% of 10,000=0.95(10,000)=9500

Test is correct for 95% of times means that 95% times drug users test results is positive and non drug users test results is negative.

Test positive for drug user= 95% of 500=0.95(500)=475

Test negative for non drug user= 95% of 500 = 0.95(9500)=9025

Test is correct for 95% of times and it means that 5% of times test is not correct. It means that 5% of times drug users test results is negative and non drug users test results is positive.

Test negative for drug user= 5% of 500=0.05(500)=25

Test positive for non drug user= 5% of 9500 = 0.05(9500)=475

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A florist gathered data about the weekly number of flower deliveries he made to homes and to businesses for several weeks. He us
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The number of deliveries that are predicted to be made to homes during a week with 50 deliveries to business is 87 deliveries

Step-by-step explanation:

The data categorization are;

The number of home deliveries = x

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The line of best fit is y = 0.555·x + 1.629

The number of deliveries that would be made to homes when 50 deliveries are made to businesses is found as follows;

We substitute y = 50 in the line of best fit to get;

50 = 0.555·x + 1.629 =

50 - 1.629 = 0.555·x

0.555·x = 48.371

x = 48.371/0.555= 87.155

Therefore, since we are dealing with deliveries, we approximate to the nearest whole number delivery which is 87 deliveries.

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The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
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