S= 70w + 750
s(19) = 70w x 19+350= 1680
so its 1680
Answer:
y = (2/5) OR y = (6/5)
Step-by-step explanation:
The first step is isolating the expression within the absolute value bars. The first thing we can do is subtract both sides by 8. If we do that, we get -2|4-5y| = -4. Now, to completely isolate the absolute value, we would have to divide by -2. This yields |4 - 5y| = 2. Finally, we can remove the absolute value bars. However, to do this, we need to first understand what an absolute value bar does to an equation. Lets say that |x| = 2. Absolute value describes the DISTANCE of some quantity from 0 (on the number line). Therefore, x (which is inside the absolute value bars) can be either positive or negative 2 (they are BOTH two units away from 0). Similarly, in this case, (4 - 5y) can either be 2 or -2 (because the absolute value of both is 2). Now we have two possible solutions to solve for:
4 - 5y = 2 OR 4 - 5y = -2
5y = 2 OR 5y = 6
y = (2/5) OR y = (6/5)
If we plug both of these answers back into the equation we can see that they both check out.
Answer:
The sample size is 
Step-by-step explanation:
From the question we are told that
The margin of error is E =4.266
The standard deviation is 
From the question we are told the confidence level is 99% , hence the level of significance is
=> 
Generally from the normal distribution table the critical value of 
is
Generally the sample size is mathematically represented as
![n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cfrac%7BZ_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%20%2A%20%20%5Csigma%20%7D%7BE%7D%20%5D%20%5E2)
=> ![n = [\frac{2.58 * 2.539 }{4.266} ] ^2](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cfrac%7B2.58%20%20%2A%20%202.539%20%20%7D%7B4.266%7D%20%5D%20%5E2)
=>
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
M(x, y) = ((x1 + x2)/2, (y1 + y2)/2) = ((-2 + 4)/2, (5 - 9)/2) = (2/2, -4/2) = (1, -2)
