2018年5月3日 · How do you use the binomial series to expand #1 / (1+x)^4#? How do you use the binomial series to expand #f(x)=(1+x)^(1/3 )#? See all questions in The Binomial Theorem

2018年3月7日 · 55 Add the first and the last numbers of the sequence and repeat for the other sets of numbers. 1 + 10 = 11 2 + 9 = 11 3 + 8 = 11 4 +7 = 11 5 + 6 = 11 This gives 5 sets of numbers that equal 11. 5 xx 11 = 55

2018年7月28日 · See proof below 1=1\\mod 5 1^{99}=1^99\\mod 5 1^{99}=1\\mod 5 Similarly, 2^{99}=2^{99}\\mod 5 3^{99}=(-2)^99\\mod 5 4^{99}=(-1)^99\\mod 5 5^{99}=0\\mod 5 \\therefore ...

2016年3月4日 · 2.576 Z score of a 99 confidence interval is 2.576. The third quartile, denoted #Q_3#, is the data value such that what percent of the values are below it?

2017年4月8日 · Sulfuric acid's molar mass is "98.08 g/mol". Step one: Find the atomic masses in each atom -- units: "g/mol" Hydrogen: 1.01 Sulfur: 32.06 Oxygen: 16 Step two: Find how many of those atoms are of the formula: Hydrogen: 2 atoms Sulfur: 1 atom Oxygen: 4 atoms Step three: Add all the atomic masses together. (Note, if there are more than one of a particular atom, you must multiply it by how many of ...

2016年8月22日 · If you add these up you get: 78.084 + 20.946 + 0.934 + 0.03 = 99.994% (the remaining material 100 - 99.994 = 0.006% consists of trace amounts of other gases) These percentages are percentage volume and one mole of gas at stand temperature and pressure occupies 22.4 liters (but that is really irrelevant to this question as the volumes cancel).

2017年10月28日 · The molar volume of #"STP"# is usually quoted at #22.4*L*mol^-1# (these definitions vary according to the syllabus you follow.... For the molar quantity of this mass of dioxygen gas we have.... #(64.0*g)/(32.0*g*mol^-1)=2*mol#.... And so to find the volume occupied by the molar quantity, we take the product....

How do you use the 68-95-99.7 rule to find the percentage of people with scores below 84? If students toss a coin 200 times each, about 68% should have proportions between what two numbers? Assume that z-scores are normally distributed with a …

2016年1月21日 · I found 31,33,35 Let us call our odd integers: 2n+1 2n+3 2n+5 and write our condition as: (2n+1)+(2n+3)+(2n+5)=99 and solve it for n: 6n+9=99 6n=90 n=90/6=15 so our numbers will be: 2n+1=31 2n+3=33 2n+5=35

2018年6月27日 · Notice that in our sequence is 0-99.4=-99.4 -99.4-99.4=0 -198.8-99.4=99.4... So we are dealing with an arithmetic sequence with difference -99.4. Each term is build up "adding" -99.4 to prior term. So the recursive formula is f(n)=f(n-1)+(-99.4)