((8 - 7.28)/7.28) x 100 = 9.89%
50%
accepted density is a part of the equation of the percent error... i.e. :experimental value- accepted value/ accepted value x100% = percent error
42?
First, understand the the percent composition of air does not change at 8000 ft. The oxygen, which is what you need to stay alive, is 21% at sea level and at 8000 ft. The density of air is what decreases as you get higher. So even thought at 10,000 feet the air is still 21% oxygen, a given breath of air, contains so few oxygen molecules that you cannot stay mentally alert. Think of a breath of air at 10,000 feet as very dilute air, with lots of empty space between the molecules. Now many jet airplanes pressurize to 8000 feet which is where that number mostly likely came from in the question. The question asks about the percent of air at 8000 feet. If we assume no humidity and we assume temperature of 70 degree ( both those things affect air density so we need to pick some levels and those are reasonable and easy for the calculation) the air density is .0554 lbm/ft3 . The density at 0 feet given the same conditions is .0745lbm/ft3 We can look at the ratio of .0554/.0745 is about 74% so the percent of "air" is about 74% at 8000 feet if we use 0 feet as 100%. If we use 30% relative humidity for both 0 and 8000 feet, the percent does not change. There is a link for calculation air density attached to this answer.
100
10 percent of 60 students is 6 students.
45.
Approximately 56% of college students in the United States are women. This percentage has been steadily increasing over the years, with women now outnumbering men in higher education enrollment.
72 is 80 percent of 90.
12.4%
easy 58 because 30+12=42 100-42=58
40 x 0.30% = 12
gaga si cherywel
10 percent of students go out instead of use electronic devices
Calculation: 28800 x 0.15 = 4320 Therefore, 15 percent of 28800 is 4320.
13 students = 1300 %
to calculate the weight from the volume, you always multiply by the density. example: If the benzene content is 2.5%v/v, then in 100mL of gasoline, you have 2.5mL of benzene. If the density of Benzene at 15.5°C is 0.81g/mL, then 2.5 x 0.81 = 2.0%w/v.