Answer:
80 ... ..........
Step-by-step explanation:
i got 100 on test
Answer:
<h2>2/3</h2>
Step-by-step explanation:
This problem bothers on probability
Given that two dice are expected to give an outcome of sum greater than 8
But in the problem statement the first die is already settled since it is showing 6 already.
Now for the second die, the possible outcomes that will add up with the first die to produce a figure greater than 8 are (3,4,5,6)
The probability of obtaining either of these numbers is 4/6= 2/3
<em>Having the first die settled, this means that the probability of obtaining a sum of 8 is 2/3</em>
<em />
#1) 4 weeks, $280
#2) (5, 38)
#3) (1, 3)
#4) Step 3
#5) (y+z=6)*-8
#6) 2x+2y=8
Explanation
#1) Setting them equal,
60x+40=50x+80
Subtract 50x from both sides:
60x+40-50x=50x+80-50x
10x+40=80
Subtract 40 from both sides:
10x+40-40=80-40
10x=40
Divide both sides by 10:
10x/10 = 40/10
x=4
Plugging this in to one of our equations,
60(4)+40=240+40=280
#2) Setting the equations equal to one another,
8x-2=9x-7
Subtract 8x from both sides:
8x-2-8x=9x-7-8x
-2=x-7
Add 7 to both sides:
-2+7=x-7+7
5=x
Plugging this in to the first equation,
8(5)-2=y
40-2=y
38=y
#3) Substituting our value from the second equation into the first one,
n-2+3n=10
Combining like terms,
4n-2=10
Add 2 to both sides:
4n-2+2=10+2
4n=12
Divide both sides by 4:
4n/4 = 12/4
n=3
Substitute this into the second equation:
m=3-2=1
#4) The mistake was made on Step 3; the 4 was left off when the equations were added.
#5) To eliminate y, we want the coefficients to be the same. To accomplish this, we will multiply the first equation by -8.
#6) In order to have infinitely many solutions, we want each coefficient as well as the constant to be a multiple of our equation. Multiplying the equation by 2, we get 2x+2y=8.
