English scores: 80, 82, 79, 84
Average of English Scores = \(\frac{80 + 82 + 79 + 84}{4}\)= \(\frac{325}{4}\) = \(81.25\)
Let the score of next English quiz be \(x\). It is given that after next quiz, the average score would be 85:
\(\frac{80 + 82 + 79 + 84 + x}{5}\)\(= 85\)
\(\frac{325 + x}{5}\)\(= 85\)
\(325 + x = 425\)
\(x = 100\)
History scores: 90, 71
Average of History Scores = \(\frac{90 + 71}{2}\) = \(\frac{161}{2}\)\(= 80.5\)
Let the score of next History quiz be \(y\). It is given that after the next quiz, the average score would be 85:
\(\frac{90 + 71 + y}{3}\)\(= 85\)
\(y + 161 = 255\)
\(y = 94\)
the sum of the scores she would need to get on her next English quiz and her next History quiz to raise each class quiz score average to 85 \(= x + y = 100 + 94 = 19\)\(4\)
The correct answer is C
Average of English Scores = \(\frac{80 + 82 + 79 + 84}{4}\)= \(\frac{325}{4}\) = \(81.25\)
Let the score of next English quiz be \(x\). It is given that after next quiz, the average score would be 85:
\(\frac{80 + 82 + 79 + 84 + x}{5}\)\(= 85\)
\(\frac{325 + x}{5}\)\(= 85\)
\(325 + x = 425\)
\(x = 100\)
History scores: 90, 71
Average of History Scores = \(\frac{90 + 71}{2}\) = \(\frac{161}{2}\)\(= 80.5\)
Let the score of next History quiz be \(y\). It is given that after the next quiz, the average score would be 85:
\(\frac{90 + 71 + y}{3}\)\(= 85\)
\(y + 161 = 255\)
\(y = 94\)
the sum of the scores she would need to get on her next English quiz and her next History quiz to raise each class quiz score average to 85 \(= x + y = 100 + 94 = 19\)\(4\)
The correct answer is C