System of equations subtraction method
In the addition/subtraction method, the two equations in the system are added or subtracted to create a new equation with only one variable. ... Substitute the variable back into one of the equations and solve for the other variable. Check the solution--it should satisfy both equations.
We have that
the expression 2f + 4f + 2 – 3-------> we can group it (2f+4f)+(2-3)
(2f+4f)+(2-3)=(6f-1)
therefore
the expression [2f + 4f + 2 – 3] is equivalent to [6f-1]
if the expression [6f-1] for f=3 is 17
then
the expression [2f + 4f + 2 – 3] for f=3 is also 17
<span>let's check it
</span>[2*3 + 4*3 + 2 – 3]--------> [6+12+2-3]=[20-3]=17------> is ok
Hey there! I am on the same one. :) I will help you out a little.
<span>Assume that all six outcomes of a six-sided number cube have the same probability. What is the theoretical probability of each roll?
• 1: 1/6
• 2: 2/6
• 3: 3/6
• 4: 4/6
• 5: 5/6
• 6: 6/6
</span>
<span>Using the uniform probability model you developed, what is the probability of rolling an even number?
1/6 Roll a number cube 25 times. Record your results here.
</span><span>
<span><span>
<span>
<span>1st
toss=</span>6</span>
</span>
<span>
<span>
<span>2nd
toss=</span>4</span>
</span>
<span>
<span>
<span>3rd
toss=</span>6</span>
</span>
<span>
<span>
<span>4th
toss=</span>6</span>
</span>
<span>
<span>
<span>5th
toss=</span>3</span>
</span>
<span>
<span>
<span>6th
toss=</span>3</span>
</span>
<span>
<span>
<span>7th
toss=</span>4</span>
</span>
<span>
<span>
<span>8th
toss=</span>2</span>
</span>
<span>
<span>
<span>9th
toss=</span>6</span>
</span>
<span>
<span>
<span>10th
toss=</span>5</span>
</span>
<span>
<span>
<span>11th
toss=</span>1</span>
</span>
<span>
<span>
<span>12th
toss=</span>4</span>
</span>
<span>
<span>
<span>13th
toss = </span>5</span>
</span>
<span>
<span>
<span>14th
toss =</span>1</span>
</span>
<span>
<span>
<span>15th
toss=</span>4</span>
</span>
<span>
<span>
<span>16th
toss=</span>2</span>
</span>
<span>
<span>
<span>17th
toss=</span>2</span>
</span>
<span>
<span>
<span>18th
toss=</span>2</span>
</span>
<span>
<span>
<span>19th
toss=</span>6</span>
</span>
<span>
<span>
<span>20th
toss=</span>5</span>
</span>
<span>
<span>
<span>21st
toss=</span>3</span>
</span>
<span>
<span>
<span>22nd
toss=</span>4</span>
</span>
<span>
<span>
<span>23rd
toss=</span>3</span>
</span>
<span>
<span>
<span>24th
toss=</span>3</span>
</span>
<span>
<span>
25
toss=5
How
many results of 1 did you have? __2____________ How
many results of 2 did you have? ____4__________ How
many results of 3 did you have? ____5__________ How
many results of 4 did you have? ______5________ How
many results of 5 did you have? ______4________
How
many results of 6 did you have? ______5________
Based
on your data, what is the experimental probability of each roll?
<span>
1. 2/25 or 0.08
2. 4/25 or 0.16
3. 5/25 or 0.24
4. 5/25 or 0.2
5.4/25 or 0.16
<span>
6. 5/25 or 0.2</span></span>Using
the probability model based on observed frequencies, what is the probability of
rolling an even number?
3/6 = ½ or 0.5
Was your experimental probability
different than your theoretical probability? Why or why not?
<span>It somewhat is! The
denominator is 25 for the experimental probability, and 6 for the theoretical
probability.</span><span>
</span><span>Have a lovely day! Cheerio. :) </span></span>
</span>
</span></span>
It was stated in the problem that a given volume of a is inversely proportional with the pressure of the system. It means that the as the volume increases, the pressure would decrease and as the volume decreases, the pressure would increase. We would express it as:
V α 1/P
To change it to an equality, we introduce a proportionality constant, k. We do as follows:
V = k/P
So, to determine what is asked, we need to first calculate for the value of k.
V = k/P
At V = 160 m^3 P = 108 cmHg
160 = k / (108)
k = 17280
Thus, at P = 87 cmHg
V = 17280 / 87
V = 198.62 m^3