Hey!
Let's write the problem.
Add like terms...
Add
to both sides.
Divide both sides by
.
Our final answer would be,
Thanks!
-TetraFish
Answer:
P = 0.006
Step-by-step explanation:
Given
n = 25 Lamps
each with mean lifetime of 50 hours and standard deviation (SD) of 4 hours
Find probability that the lamp will be burning at end of 1300 hours period.
As we are not given that exact lamp, it means we have to find the probability where any of the lamp burning at the end of 1300 hours, So we have
Suppose i represents lamps
P (∑i from 1 to 25 (
> 1300)) = 1300
= P(
> 
) where represents mean time of a single lamp
= P (Z> 
) Z is the standard normal distribution which can be found by using the formula
Z = Mean Time () - Life time of each Lamp (50 hours)/ (SD/
)
Z = (52-50)/(4/
) = 2.5
Now, P(Z>2.5) = 0.006 using the standard normal distribution table
Probability that a lamp will be burning at the end of 1300 hours period is 0.006
I’m not sure but this could help
A is located, as you look A is past -2 almost hitting -3 making the mark 2.8 so the answer is A
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
