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( $9$ ) Jace mixed $17$ gallons of two fruit drink brands. Brand A contains $14 \%$ fruit juice and Brand $B$ contains $48 \%$ fruit juice. If the resulting mixture contains $30 \%$ fruit juice, how many gallons of each should he use? $\newline$ Let $\qquad$ $=$ $\qquad$ and Let $\qquad$ $\square+L=\square$ $\newline$ ( $13$ ) ( $14$ )

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Experimental probability

Full solution

Q. (

9

) Jace mixed

17

gallons of two fruit drink brands. Brand A contains

14 \%

fruit juice and Brand

B

contains

48 \%

fruit juice. If the resulting mixture contains

30 \%

fruit juice, how many gallons of each should he use?

\newline

Let

\qquad

=

\qquad

and Let

\qquad

\square+L=\square

\newline

(

13

) (

14

)

Define Variables: Let $x$ be the gallons of Brand A and $y$ be the gallons of Brand B. We know that $x + y = 17$ gallons.
Set up Juice Content Equation: Set up the equation for the juice content: $0.14x$ (juice from Brand A) + $0.48y$ (juice from Brand B) = $0.30 \times 17$ (total juice in the mixture).
Substitute Variables: Substitute $y = 17 - x$ into the juice content equation: $0.14x + 0.48(17 - x) = 5.1$ .
Simplify and Solve: Simplify and solve for $x$ : $0.14x + 8.16 - 0.48x = 5.1$ , $-0.34x + 8.16 = 5.1$ , $-0.34x = -3.06$ , $x = 9$ gallons.

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