Replace with 16 − 2 and simplify. 1 , y = 16−2,2, (4,2), (−4,k), k, x-intercepts
Answer:
Check the explanation
Step-by-step explanation:
Ans=
A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134
M = 10: 1 – 0.9973^10 = 0.0267
M = 20: 1 – 0.9973^20 = 0.0526
M = 30: 1 – 0.9973^30 = 0.0779
M = 50: 1 – 0.9973^50 = 0.126
18)
Ans=
Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values
You can solve for the velocity and position functions by integrating using the fundamental theorem of calculus:
<em>a(t)</em> = 40 ft/s²
<em>v(t)</em> = <em>v </em>(0) + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
<em>v(t)</em> = -20 ft/s + ∫₀ᵗ (40 ft/s²) d<em>u</em>
<em>v(t)</em> = -20 ft/s + (40 ft/s²) <em>t</em>
<em />
<em>s(t)</em> = <em>s </em>(0) + ∫₀ᵗ <em>v(u)</em> d<em>u</em>
<em>s(t)</em> = 10 ft + ∫₀ᵗ (-20 ft/s + (40 ft/s²) <em>u</em> ) d<em>u</em>
<em>s(t)</em> = 10 ft + (-20 ft/s) <em>t</em> + 1/2 (40 ft/s²) <em>t</em> ²
<em>s(t)</em> = 10 ft - (20 ft/s) <em>t</em> + (20 ft/s²) <em>t</em> ²
Answer:
2 times 5 times 5 times 5 times 5
