What is the gcf of 6 and 18

The numbers on a football field indicate 10 yard increments. you walk around the perimeter of a football field between the pylons. you walk a distance of 30623 yards. find the area and perimeter of the indicated regions. write your answers as mixed numbers, if necessary. a. one end zone: perimeter: yd area: yd2 b. the playing field (not including end zones): perimeter: yd area: yd2 c. the playing field (including end zones): perimeter: yd area: yd2

In this problem, we explore the effect on the standard deviation of multiplying each data value in a data set by the same constant. consider the data set 14, 6, 8, 15, 15. (a) use the defining formula, the computation formula, or a calculator to compute s. (round your answer to one decimal place.) s = 4.28 (b) multiply each data value by 3 to obtain the new data set 42, 18, 24, 45, 45. compute s. (round your answer to one decimal place.) s = 12.83 (c) compare the results of parts (a) and (b). in general, how does the standard deviation change if each data value is multiplied by a constant c? multiplying each data value by the same constant c results in the standard deviation remaining the same. multiplying each data value by the same constant c results in the standard deviation being |c| times as large. multiplying each data value by the same constant c results in the standard deviation increasing by c units. multiplying each data value by the same constant c results in the standard deviation being |c| times smaller. (d) you recorded the weekly distances you bicycled in miles and computed the standard deviation to be s = 3.8 miles. your friend wants to know the standard deviation in kilometers. do you need to redo all the calculations? yes no given 1 mile ≠1.6 kilometers, what is the standard deviation in kilometers? (enter your answer to two decimal places.)

Anull hypothesis states that the difference between 8 hours of sleep and 3 hours of sleep has no effect on the number of pictures a student will remember during a picture memory interface test (pmit). examining the mit database, an extremely small t-value of 0.1611 is found. this t-value is much smaller than the smallest t-value on the chart for p-value reference. from this information we can: select one: a. accept the null hypothesis because the p-value obtained shows that the difference between the two groups being tested is not statistically significant b. accept the null hypothesis because the p-value obtained was statistically significant c. reject the null hypothesis because of the p-value obtained d. reject the null hypothesis because the data obtained is statistically significant