Agroup of educators want to determine how effective tutoring is in raising students' grades in a math class, so they arrange free tutoring for those who want it. then they compare final exam grades for the group that took advantage of the tutoring and the group that did not. suppose the group participating in the tutoring tended to receive higher grades on the exam. does that show that the tutoring worked? if not, explain why not and suggest a confounding variable.

Step-by-step explanation:

Given that a group of educators want to determine how effective tutoring is in raising students' grades in a math class, so they arrange free tutoring for those who want it.   Whenever we want to compare the two groups the variance of the two groups also is very important to arrive at the test statistic and coming to conclusion. Here no mention is made about variance or standard deviation of the two groups.  Also there may be confounding variables present such as the time devoted by the students for study, the intelligent quotient of the students, the interest of the students in the subject, previous scores.

here is your mark brainliest

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%