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Dimas [21]
4 years ago
6

When Megan solved this system using the elimination method , she added the equations together and got the equation 4y = 16. Then

she solved that equatio n and got y = 4. Then she substituted 4 in the first equation for y and solved for x. She got x = 4. The point (4, 4) is not the correct solution. Please state in a complete sentence where she made her first mistake. What is the correct solution to the system ? Show your work in the space provided. 2x + 3y = 20 - 2x + y = 4
Mathematics
1 answer:
weqwewe [10]4 years ago
8 0

Answer:

Megan added the equations together but  she subtract 4 instead of addition of 4. The correct solution is (1,6).

Step-by-step explanation:

The given equations are

2x+3y=20        ...(1)

-2x+y=4            .... (2)

Add both equations to eliminate x.

2x+3y-2x+y=20+4

4y=24

Megan make a mistake in this step. She added the equations together but she subtract 4 instead of addition of 4.

Divide both sides by 4.

y=6

Put this value in equation (1).

2x+3(6)=20

2x+18=20

2x=2

Divide both sides by 2.

x=1

Therefore the correct solution is (1,6).

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Nonamiya [84]

Answer:

P(X>7.3) = \int_{5.5}^{7.3} \frac{1}{2} dx = \frac{1}{2} x \Big|_{6.5}^{7.3} = \frac{1}{2} (7.3-6.5)= 0.4

And then we can find the probability replacing the last result and we got:

P(X>7.3)= 1- P(X

Step-by-step explanation:

For this case we have the following density function given:

f(x) = \frac{1}{2} , 5.5 \leq X \leq 8.5

And we want to find the following probability:

P(X>7.3)

We can find this probability with the complement rule like this:

P(X>7.3)= 1- P(X

And we can find the following probability:

P(X>7.3) = \int_{5.5}^{7.3} \frac{1}{2} dx = \frac{1}{2} x \Big|_{6.5}^{7.3} = \frac{1}{2} (7.3-6.5)= 0.4

And then we can find the probability replacing the last result and we got:

P(X>7.3)= 1- P(X

7 0
3 years ago
The cost of 300 metres optic fibre is $200 what will be the cost if 1 metre of optic fibre.
Elden [556K]

The answer is:  " $0.67 / metre " ;  
                        or, write as:  " [ \frac{2}{3} dollar] per metre."
____________________

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

This is a unit rate problem:
   Find the cost in dollars per [single unit—in this case: "metre(s)", "<em>m</em>" ;
_________________
 \frac{200 dollars}{300m} = \frac{? dollars}{1m} ;  Solve for the "?" <u>amount of dollars</u>.
_________________
Method 1)  "Cross-factor multiply" :
<u>Note</u>:  <u>Given</u><u>:</u> " \frac{a}{b} = \frac{c}{d} ;  [b\neq 0;  d\neq 0] " ;

 then:   ⟷  " ad = bc " ;
_________________

So:  " \frac{200 dollars}{300m} = \frac{? dollars}{1m} " ;

  then:   ⟷  " (200*1) = (300* ?) ;

     → "  (200) = (300  * ?)" ;

     ↔ " (300 * ?) = 200 " ;  

Let "x" refer to the "?" ; the "<u>unknown value</u>" (<u>in dollars</u>);

                for which we are trying to solve.

⇒ " 300x = 200 " ;  Solve for "x" ;
     → Divide each side of the equation by: "300" ;

       to isolate "x" on the "left-hand side" of the equation; & to              solve  for "x" ;
   ⇒  300x/ 300 = 200/300 ;

    To get:  " x = \frac{2}{3} dollars" ;  

= $0.666666666... ;

round to: $0.67 .

{since: " 1 dollar = $1.00 = 100 cents" ;
and since: "\frac{2}{3} " ;

= ["2 ÷ 3" = 0.666666...." ] ;  

⇒  round to:  "0.67 " ;  

and write as:
__________

⇒  $0.67 / metre ;   or, write as:
    " [ \frac{2}{3} dollar] per metre."
_________

Method 2)  <u>Note</u>:
\frac{200 dollars}{300m} = \frac{? dollars}{1m} ;

⇔   = "  \frac{200}{300}  \frac{dollars}{metre} " ;
      {<u>Note</u>:  "[200÷300]" can be simplified by "canceling out" the two (2) zeros in both the numerator and the denominator.}.

   = " \frac{2 dollars}{3m} ";
_________

  =  i.e. " \frac{\frac{2}{3} dollars}{metre} " or;  \frac{0.67 dollars} {m} ;  or:  <u>$0.67 / metre</u> .

         ⇒  {since:  " \frac{2}{3} of 1 dollar" = " \frac{2}{3} dollars of 100 cents" ;
        ⇒  {since:  "1 dollar = 100 cents"} ;

           ⇒ {and: " \frac{2}{3} " = 0.6666666666666... " } ;  

 → round to:  "0.67 " ; and thus: $0.67 / metre.
_________
Note that using either of these 2 (two) methods result in the same answer.
               Hope this is helpful!  Best wishes to you within your academic pursuits!
             _________________

7 0
3 years ago
Shirley has a credit card that uses the previous balance method. The opening
vesna_86 [32]

Answer:

= $44.19

Step-by-step explanation:

APR = 19%

Billing cycle = 30 days

Balance = $2830

Let's first find the daily rate,

Daily rate =  \frac{0.19}{365} = 0.000521

The daily rate = 0.000521

To calculate the amount Shirley was charged interest for the billing cycle, we use:

Daily rate * billing cycle * balance

Where,

Daily rate

=  \frac{0.19}{365}

Billing cycle = 30 days

Balance = $2830

Therefore, expression to be used =

(\frac{0.19}{365}  * 30) * 2830 \\= $44.19

8 0
3 years ago
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b ≈ 101.5

6 0
3 years ago
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