The minimum distance is the same from both points because their lengths are equal.
A normal distribution is most effective when dealing with a substantial sample size. Without knowing how many containers there are, it's challenging to determine if it’s suitable for modeling the container weights.
Solution
To find the angle m∠p.
Method of proof
In triangle ΔDAB, which is a right triangle
Applying the Pythagorean theorem gives us
Hypotenuse² = Perpendicular² + Base²
DB² = AB² + AD²
Where AB = 5 units
AD = 6 units
Substituting in the formula outcomes in
DB² = 5² + 6²
= 25 + 36
= 61
= approximately 7.8 units
The triangle ΔDCB is also right-angled.
Using the trigonometric identity here.
Given that DC = 4 units and DB ≈ 7.8 units,
Substituting these values into the trigonometric identity gives us.
Thus, we find that ∠p ≈ 59.15 °
For this scenario, we start with the function:
<span>w (x) = - 5 (x-8) (x + 4)
</span><span>Reorganizing yields:
</span><span>w (x) = - 5 (x² + 4x - 8x - 32)
</span><span>w (x) = - 5x² - 20x + 40x + 160
</span><span>w (x) = - 5x² + 20x + 160
</span><span>Next, we take the derivative:
</span><span>w '(x) = - 10x + 20
</span><span>Setting this to zero and solving for x:
</span><span>0 = -10x + 20
</span><span>10x = 20
</span><span>x = 20/10
</span><span>x = 2 seconds
</span><span>Substituting back:
</span><span>w (2) = - 5 (2-8) (2 + 4)
</span><span>w (2) = - 5 (-6) (6)
</span><span>w (2) = 180 meters
</span>Conclusion:
The peak height attained by the stone is:
w (2) = 180 meters
To resolve this kind of question, we need to analyze it on a yearly basis.
At the end of the first year, the tree would experience a growth of 2% = 102% of its initial height.
Year 1: 102% x 50 = 51 feet
Year 2: 103% x 51 = 52.02 feet
I hope this was helpful!! xx
