How can thinking like a scientist and having knowledge of scientific experiments and reasoning make you a wiser consumer?

Use the frequency histogram to complete the following parts. ​(a) identify the class with the​ greatest, and the class with the​ least, relative frequency. ​(b) estimate the greatest and least relative frequencies. ​(c) describe any patterns with the data. female fibula lengths 30.5 31.5 32.5 33.5 34.5 35.5 36.5 37.5 38.5 39.5 0 0.05 0.1 0.15 0.2 0.25 length (in centimeters) relative frequency a histogram titled "female fibula lengths" has a horizontal axis labeled "length in centimeters" from 30.5 to 39.5 in increments of 1 and a vertical axis labeled "relative frequency" from 0 to 0.25 in increments of 0.05. the histogram contains vertical bars of width 1, where one vertical bar is centered over each of the horizontal axis tick marks. the approximate heights of the vertical bars are listed as follows, where the label is listed first and the approximate height is listed second: 30.5, 0.02; 31.5, 0.04; 32.5, 0.05; 33.5, 0.13; 34.5, 0.22; 35.5, 0.25; 36.5, 0.13; 37.5, 0.06; 38.5, 0.09; 39.5, 0.01. ​(a) the class with the greatest relative frequency is nothing to nothing centimeters. ​(type integers or decimals. do not round. use ascending​ order.)

Find the equation for the plane through upper p 0 left parenthesis negative 4 comma negative 8 comma negative 5 right parenthesis perpendicular to the following line. xequalsnegative 4 minus t, yequalsnegative 8 plus 2 t, zequals3 t, minusinfinityless thantless thaninfinity

Assume you measured the mass of the cart to be (500 ± 1) g and the mass of the additional mass you put on the cart to be (500 ± 1) g as well. since the scale you are using in the lab cannot measure objects heavier than 600g you will have to sum up individual pieces and propagate the error. so what would be the mass and the standard error of the cart and the mass