When we are given a system of 3 linear equations, with 3 variables, we proceed as follows:
We consider 2 pairs or equations, for example (1, 2) and (2, 3), and eliminate one of the variables in each pair, creating a system of 2 linear equations with 2 unknowns.
Note that the third equation contains -2y which can be used to eliminate easily -6y in the second equation, and -4y in the fourth.
i) consider equations 1 and 3:
-3x-4y-3z=-7
5x-2y+5z=9
multiply the second equation by -2:
-3x-4y-3z=-7
-10x+4y-10z=-18
adding the 2 equations we have -13x-13z=-25
ii) consider equations 2 and 3. Multiply the third equation by -3:
2x-6y+2z=3
-15x+6y-15z=-27
adding the 2 equations we have -13x-13z=-24
So we got -13x-13z is -25, but also -24. this means the system is inconsistent, so it has no solution.
Answer: the system has no solutions
Hello There!
You want to know how many cookies can be made using 5 1/2 cups of flour.
1 x 2 2
-------------- = -----
2 x 11 22
2/22 simplified is
1/11 So, You can make
11 cookies with 5 1/2 cups of flour.
Hope this helped!
Answer:
2923.34 in^3
Step-by-step explanation:
Volume = Pi x (r) ^2 x height
PI =3.14,
, R = 7inch (d =radius /2 >> 14/2)
H =19 inch
Volume = 3.14 x 49in^2 x 19 inch
= 2923.34 in^3
Answer:
3
Step-by-step explanation:
3×3=9+2=11. so 3 is the answer.
Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
