Answer:
Given:
n = 15
c = 95% = 0.95
Let us assume:
α = 0.05
Determine the difference in value of each pair.
→See attached for the table←
→See attached for the for the workings←
Result
a) There is sufficient evidence to reject the claim that the two methods provide the same mean value for natural vibration frequency.
b) (-10.9646, -0.0068)
Answer:
1 ≥ t ≤ 3
Step-by-step explanation:
Given
h(t) = -16t² + 64t + 4
Required
Determine the interval which the bar is at a height greater than or equal to 52ft
This implies that
h(t) ≥ 52
Substitute -16t² + 64t + 4 for h(t)
-16t² + 64t + 4 ≥ 52
Collect like terms
-16t² + 64t + 4 - 52 ≥ 0
-16t² + 64t - 48 ≥ 0
Divide through by 16
-t² + 4t - 3 ≥ 0
Multiply through by -1
t² - 4t + 3 ≤ 0
t² - 3t - t + 3 ≤ 0
t(t - 3) -1(t - 3) ≤ 0
(t - 1)(t - 3) ≤ 0
t - 1 ≤ 0 or t - 3 ≤ 0
t ≤ 1 or t ≤ 3
Rewrite as:
1 ≥ t or t ≤ 3
Combine inequality
1 ≥ t ≤ 3
Step-by-step explanation:
x² - 2x - 15 = 0
x1 = 5
x2 = -3
Answer:
3/4
Step-by-step explanation:
Answer:
Black Screen so sorry
Step-by-step explanation:
