Answer:
Follows are the solution to the given points:
Step-by-step explanation:
In point a:
In this sense, describe what type of error I will be.
Type I error: to conclude that perhaps the mean bulb life would be less than three hours when it becomes (at least) 3 hours.
In point b:
Describe throughout this context what the Type II error becomes.
An error of type II: never assuming that its bulbs' mean lifetime is much less than 3 hours. three hours at least
In point c:
What error — type I and type II — would further impact the interaction between the manufacturer and the customer?
A Type II error is probably further problematic because it means that even the buyer will buy bulbs that do not last long.
Given
d(t)=-16t^2-4t+382
1. times at which rock is 310 from ground
solve d(t)=310=-16t^2-4t+382
=>
16t^2+4t-72=0
t=(-4+/-sqrt(4^2+4*16*72)/(2*16)
=> t=-9/4 or t=2
Reject negative root from context, so
time=2 seconds.
2.
maximum is when d'(t)=0, or
-32t-4=0 => t=-1/8
Therefore rock was descending since t=0 (can also see from -4t, i.e. initial velocity is -4 ft/s)
Rock reaches ground when d(t)=0, i.e.
-16t^2-4t+382=0
t=-5.01 or t=4.763 seconds. reject negative root.
Therefore, domain, t=0 to 4.763 seconds
range = 0 to 382 ft
<span>f(i) = i³ - 2(i)²
Use the facts, i³ = -i and i² = -1
f(i) = 2 - i</span>
Answer:
70
Step-by-step explanation:
hope this helps!