Question

12. At the baseball stadium, four hot dogs and

Answer 1

Answer:

The cost of one hot dog can't be determine from the given condition

Step-by-step explanation:

From the question,

Total cost of four hot dogs and three bags of candy is $14.37 and twelve hot dogs and nine bags of candy is $24.74.

So,

Let the number of hot dog be x

and the number of bags of candy be y.

According to question,

Equations formed by given statement are,

                     $4{x} + 3{y} = 14.37$                     .................(1)

                     $12{x} + 9{y} = 24.74$                   .................(2)

Here, these are the linear equation in two variables says x & y.

                     $a1=4$      $b1=3$     $c1=14.37$

                     $a2=12$    $b2=9$    $c1=24.74$

When ,

            $\frac{a1}{a2}$  ≠  $\frac{b1}{b2}$  = $\frac{c1}{c2}$             gives Unique solution.

       

           $\frac{a1}{a2}$  =  $\frac{b1}{b2}$  =  $\frac{c1}{c2}$              gives Infinite many solutions.

          $\frac{a1}{a2}$  =  $\frac{b1}{b2}$   ≠  $\frac{c1}{c2}$              gives No solution.

From Equation (1) & (2) we get ,

         $\frac{4}{12}$   =  $\frac{3}{9}$   ≠  $\frac{14.37}{24.37}$

This condition says that the given equation have no solution.

Hence the cost of one hot dog can't be determine from the given condition.