Answer:
position 1 5 8 12 19 25
term -8 8 20 36 64 88
Step-by-step explanation:
(n - 1) is position ( n ∈ N)
d is the distance between the numbers in the sequence
a1 is the first number in the sequence
we have the fuction: a(n) = a1 + (n - 1)d
see in the table, with position = 1, term = -8 => a1 + d = -8
position = 25, term = 88 => a1 + 25d = 88
=> we have: a1 + d = -8
a1 + 25d = 88
=> a1 = -12
d = 4
=> a(n) = -12 + 4(n - 1)
=> term = 8, position = (8 + 12)/4 = 5
position = 8, term = -12 + 4.8 = 20
term = 36, position = (36 + 12)/4 = 12
position = 19, term = -12 + 4.19 = 64
Emily drove 3/7 of the total distance and holly drove the rest which explains that holly drove 4/7 of the whole distance.
if 4 parts of the distance equals 24 miles, then 1 part would be 24/4 is 6.
since emily drove 3 parts of the whole distance, it is 6 × 3 = 18 miles.
travelling time of emily = 18/45 = 0.4 hr
travelling time of holly = 24/54 = 0.4 hr
--> total traveling time = 0.8 hr
i hope it is right
The answer is C
It’s C because like terms have the same variable.
Answer:
a) 83%
b) 0.892
Step-by-step explanation:
percentage that attends class on friday = 74%
percentage that pass because they attend class on friday = 88%
percentage that pass but did not go to school on friday = 20%
a) percentage of students expected to pass the course
= (74% x 88%) +(88% x 20%)
= 0.6512 + 0.176
= 0.8272
= 83%
b) If a person passes the course, what is the probability that he/she attended classes on Fridays
= 74% divided by 83%
= 0.892
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)
moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g
Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05
N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
