6. the density of water is 1 g/ml. in an experiment, a student performs 3 trials and calculates the density…

Explanation:

Step1: Calculate accuracy

Accuracy measures how close a measurement is to the true - value. The true density of water $\rho_{true}=1\ g/mL$. Let the measured values be $\rho_1 = 1.5\ g/mL$, $\rho_2=1.83\ g/mL$, $\rho_3 = 0.89\ g/mL$. The average of the measured values $\bar{\rho}=\frac{1.5 + 1.83+0.89}{3}=\frac{4.22}{3}\approx1.41\ g/mL$. The accuracy error $E_{a}=\vert\bar{\rho}-\rho_{true}\vert=\vert1.41 - 1\vert=0.41\ g/mL$. Accuracy (as a percentage) $A=\frac{\vert\bar{\rho}-\rho_{true}\vert}{\rho_{true}}\times100=\frac{0.41}{1}\times100 = 41%$. In h/10, accuracy $=4.1$ (where we assume h represents 100% and we want the value out of 10).

Step2: Calculate precision

Precision is related to the spread of the data points. First, find the range of the data. The range $R=\max(\rho_1,\rho_2,\rho_3)-\min(\rho_1,\rho_2,\rho_3)$. $\max(1.5,1.83,0.89)=1.83$ and $\min(1.5,1.83,0.89)=0.89$. $R = 1.83 - 0.89=0.94\ g/mL$. Precision (as a measure related to the range out of 10 for a scale of 0 - 100% precision) is calculated as follows: A smaller range means higher precision. If we assume a linear relationship between range and precision on a 0 - 10 scale, and a range of 0 means 10 (perfect precision) and a large range means 0 precision. Let's assume a simple linear formula $P = 10-\frac{R}{ \text{max possible range}}\times10$. Here, if we assume the max - possible range for density values around 1 g/mL could be say 2 g/mL (from 0 to 2), then $P = 10-\frac{0.94}{2}\times10=10 - 4.7 = 5.3$.

Answer:

Accuracy: 4.1 h/10; Precision: 5.3 h/10