Question

there were 75 students in a precalculus class. every student in the class had a calculator. every calculator was either a tasio or an arkansas instruments (ai). there were nine more ais than tasios. four times as many students had graphing calculators as had non - graphing calculators. ten students had non - graphing ais. how many students had graphing tasios?

Explanation:

Step1: Set up equations

Let \( T \) be the number of Tasios and \( AI \) be the number of Arkansas Instruments. We know \( AI=T + 9\) and \( T+AI=75\). Substitute \( AI=T + 9\) into \( T+AI=75\):
\( T+(T + 9)=75\)
\( 2T+9 = 75\)
\( 2T=75 - 9=66\)
\( T = 33\), so \( AI=33 + 9=42\)

Let \( x \) be the number of non - graphing calculators and \( y\) be the number of graphing calculators. We know \( y = 4x\) and \( x + y=75\). Substitute \( y = 4x\) into \( x + y=75\):
\( x+4x=75\)
\( 5x=75\)
\( x = 15\), \( y=60\)

Step2: Find the number of graphing Tasios

We know the number of non - graphing \( AI\) is \( 10\). The number of non - graphing calculators \( x = 15\), so the number of non - graphing Tasios is \( 15-10 = 5\)
The number of Tasios \( T = 33\). Let \( G_T\) be the number of graphing Tasios. Then \( G_T=T-(15 - 10)\)

\( G_T=33-5=28\)

Answer:

28