Answer:
We conclude the equation is linear because it can be rewritten in the form 
.
Hence, option D is correct.
Step-by-step explanation:
The slope-intercept form of the line or linear equation
where
is the slope
is the y-intercept
<u>Important Tip:</u>
The graph of a linear equation is always a straight line.
Convert the given equation in the slope-intercept form
subtract 18x from both sides
simplify
divide both sides by 9
Now, comparing the equation with a slop-intercept form of linear equation
- The y-intercept b = -416/9
Therefore, we conclude the equation is linear because it can be rewritten in the form .
From the attached graph, is also clear that the graph of the equation is a straight line.
Hence, option D is correct.
a) 4x +6
Add up all the sides to calculate perimeter
Answer:
74.36 g of aluminium acetate.
730.27g of aluminium acetate.
- to the nearest hundredth.
Step-by-step explanation:
Acetic acid is usually written as CH3COOH.
a. 6CH3COOH + Al(OH)3 ---> Al(CH3COO)3 + 9H2O
So 6 moles of acetic acid produce 1 mole of aluminium acetate.
Using the molecular masses
6*( 1.008*4 + 12.011*2 + 16 *2) g acetic acid gives (26.98+3(36.032+ 2*12.011)
348.228 g acetic acid gives 207.142 g Al acetate.
So 125 g gives (207.142 / 348.228) * 125
= 74.36 g of aluminium acetate.
b.
(26.98 + 3*16 + 3 * 1.008) g of Al(OH)3 gives 207.142 g Al acetate
78.004 g gives 207.142 g Al acetate
275 g gives (207.142 / 78.004) * 275
= 730.27g Al acetate.
Answer:
c
Step-by-step explanation:
Answer:
a) No
b) 42%
c) 8%
d) X 0 1 2
P(X) 42% 50% 8%
e) 0.62
Step-by-step explanation:
a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.
b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6
P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7
P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
c) P(win first game) = 0.4
P(win second game) = 0.2
P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%
d) X 0 1 2
P(X) 42% 50% 8%
P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%
e) The expected value 
f) Variance 
Standard deviation 
