<span>Raul only hit 19 times out of 45 bats.
Now, let’s find the average decimal value of Raul’s hit.
=> 19 hits
=> 45 bats
x = value in decimals
=> 19 = x * 45
=> 19 = 45x
Now, divide both sides with 45
=> 19/45 = 45x / 45
=> 0.4222222 = x
x = 0.422222……..
Since it is a repeating decimals, and we only need to find the thousandths
value. Let’s get the value of Thousandths.
=> 4 = tenths
=> 2 = hundredths
=> 2 thousandths
Therefore the value of 19 hits out of 45 bats in decimals is
=> .422 </span>
Answer:
The Answer Is B
Step-by-step explanation:
-4/5 = -1 (Rounded)
3/5 = 1/2 (Rounded)
-6/7 = -1 (Rounded)
5/6 = 1 (Rounded)
If your looking for the whole thing it would be:
(-1)[1/2](-1)(1) = x
Your Welcome! <3
A chemical a element js a specific kind of atom
Answer:
(a) The average grade point is 2.5.
(b) The relative frequency table is show below.
(c) The mean of the relative frequency distribution is 0.3333.
Step-by-step explanation:
The given data set is
4, 4, 4, 3, 3, 3, 1, 1, 1, 1
(a)
The average grade point is
Therefore the average grade point is 2.5.
(b)
The relative frequency table is show below:
x f Relative frequency
4 3 
3 3
1 4 
(c)
Mean of the relative frequency distribution is
Therefore the mean of the relative frequency distribution is 0.3333.
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
