From the system of equations, we have 3 unknowns and 3 equations. Therefore, it can be solved.
<span>3x - 2y + 2z = 30 (1)
</span><span>-x + 3y - 4z = -33 (2)
</span><span>2x - 4y + 3z = 42 (3)
We can write equation 2 x as a function of y and z and substitute the new equation to the other equations.
</span>x =3y - 4z + 33 (2)
3 (3y - 4z + 3)<span> - 2y + 2z = 30 (1)
</span>2 (3y - 4z + <span>3)</span> - 4y + 3z = 42 (3)
We simplify and solve the equations and we get values,
x = 4
y = -7
z = 2
Answer:
(0, 207)
Step-by-step explanation:
We calculate y-intercept by substituting 0 for x.
The equation becomes:
3*62-6*(0)+21=207
y intercept is (0, 207)
Step-by-step answer:
Given:
mean, mu = 200 m
standard deviation, sigma = 30 m
sample size, N = 5
Maximum deviation for no damage, D = 100 m
Solution:
Z-score for maximum deviation
= (D-mu)/sigma
= (100-200)/30
= -10/3
From normal distribution tables, the probability of right tail with
Z= - 10/3
is 0.9995709, which represents the probability that the parachute will open at 100m or more.
Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.
P(all five safe) = 0.9995709^5 = 0.9978565
So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m
