Start by reviewing your knowledge of natural logarithms. If we take the ln of both sides we get e^z=ln(1). Do the same thing again and wheel about the ln(ln(1)). There's going to be complex solutions, Wolfram Alpah gets them but let me know if you figure out how to do it?
Answer:
(x,y,z) -> (2,4,1)
Step-by-step explanation:
-7x + y + z = -9
-7x + 5y - 9z = -3
7x - 6y + 4z = -6
Pick two pairs:
-7x + 5y - 9z = -3
7x - 6y + 4z = -6
and
-7x + y + z = -9
-7x + 5y - 9z = -3
Eliminate the same variable from each system:
-7x + 5y - 9z = -3
7x - 6y + 4z = -6
+ 5y - 9z = -3
- 6y + 4z = -6
<u><em>-1y - 5z = - 9</em></u>
-7x + y + z = -9
-7x + 5y - 9z = -3
-7x + y + z = -9
7x - 5y + 9z = 3
<u><em>-4y - 10z = -6</em></u>
Solve the system of the two new equations:
-1y - 5z = - 9 -> -4 ( -1y - 5z = - 9) -> 4y + 20z = 36
-4y - 10z = -6 -> -4y - 10z = -6 -> -4y - 10z = -6
10z = 30
Thus, z = 3
-4y - 10z = -6
-4y - 10(3) = -6
-4y - 30 = -6
-4y = 24
Thus, y = -6
Substitute into one of the original equations:
-7x + y + z = -9
-7x + (-6) + (3) = -9
7x + -3 = -9
7x = -6
x =
Complete question :
The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a standard deviation of .29.Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.
Answer:
0.10868
Step-by-step explanation:
Given that :
Mean (m) = 3.02
Standard deviation (s) = 0.29
Sample size (n) = 20
Probability of 3.10 GPA or higher
P(x ≥ 3.10)
Applying the relation to obtain the standardized score (Z) :
Z = (x - m) / s /√n
Z = (3.10 - 3.02) / 0.29 / √20
Z = 0.08 / 0.0648459
Z = 1.2336940
p(Z ≥ 1.2336) = 0.10868 ( Z probability calculator)
Answer:
80
Step-by-step explanation:
