Answer:
0.02566
0.972559
Step-by-step explanation:
Given that :
Sample size (n) = 36
σ² = 6
sample variance S^2
a) greater than 9.1;
b) between 3.5 and 10.5.
The degree do freedom (df) = n - 1
S² > 9.1
P(S² > 9.1) = X² > ((n - 1) * S²) / 6
P(S² > 9.1) = ((36 - 1) * 9.1) / 6
P(S² > 9.1) = (35 * 9.1) / 6
P(S² > 9.1) = 53.083
P(X² > 53.08) = 0.02566 ( chi square distribution calculator)
b) between 3.5 and 10.5.
((36 - 1) * 3.5) / 6 ≤ S² ≤ ((36 - 1) * 10.5) / 6
20.416666 ≤ S² ≤ 61.25
Using the Chisquare distribution calculator :
P(X² > 20.42) = 0.9765
P(X² > 61.25) = 0.003941
Hence,
0.9765 - 0.003941 = 0.972559
Answer:
54/5
Step-by-step explanation:
(3/5)(18)=54/5
Julia’s score is <em>y</em> and Tina’s score is <em>y</em> + 6.
Step-by-step explanation:
Let us take Julia’s score be <em>y</em>.
First Tina scored 4 points more than Julia, so the expression is <em>y</em> + 4.
Then, Tina earned 2 points as extra credic, now the expression becomes
(y + 4) + 2 = y + 6.
Hence Julia’s score is <em>y </em>and Tina’s score is <em>y</em> + 6.
Dy/dx=-.08x^3+.72x
d2y/dx2=-.24x^2+.72
.08x(9-x^2) so we have extrema at x=-3,0,3
d2y/dx2 (-3)=-1.44, (0)=.72, (3)=-1.44
So you have two global maximums at (±3,0)
We have a local minimum at (0,-1.62)
There are no global minimums as the function decreases without bound as x approaches ±oo