Answer:
18
Step-by-step explanation:
70^2 = 4900 < 5607 < 6400 = 80^2
so sqrt(5607) is a two-digit number with tenth-digit 7
71^2 = 5041, 72^2 = 5184, 73^2 = 5329, 74^2 = 5476, 75^2 = 5625
so the smallest square bigger than 5607 is 75^2, which is 5625
so the number should ne 5625 - 5607 = 18
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.
The circumference of a circle is determined by the formula 2πr.
I'd say divide the circumference by two and youll be left with 60.5=πr .
then divide 60.5 by 3.14 to get r
which is approximately 19.26
now use the circle formula for area
A=πr²
A in terms of pi is
A = π*(19.26)²
The empirical probability of a 4 or 5 is
(count of outcome = 4 or 5)/(count of all outcomes) = (44 +40)/240 = 7/20
The problem statement asks you to assume the outcomes will remain in this proportion going forward. That is, the number of outcomes (n) of 4 or 5 in the next 60 rolls of the number cube is expected to satisfy
7/20 = n/60
Multiplying that by 60 gives
n = 60*(7/20) = 21
Selection D is appropriate.
