A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i. e., accurate to +/-1 mm) at 20.0°C. If she measures the distance between two stakes to be 85.175 m on a 10.0°C day, does she need to add or subtract a correction factor to get the true distance? How large in mm is the correction factor? The coefficient of linear expansion of steel is 1.1 times 10^(-5) per Kelvin.