Answer:
B. It is reasonable to use the? z-interval procedure in this case? since, although the sample is small? (size less than? 15), the variable under consideration is very close to being normally distributed.
Step-by-step explanation:
answer b is considered to be correct because we know that the population is normal and the standard deviation is known, which allows using the interval z, the answer A is not correct because although the option to use the interval z is given, which is correct, the large sample is not favored, the answer C and D are incorrect because they both reject the use of the z interval and in d it is further rejected that although there is a normal distribution the sample is not, which is false
60min or 1 hour
Because 2km is equal to 2000m
Therefore you multiply 12 min by 5 (because 2000m x 5 is 10000m)
Answer:
See Explanation
Step-by-step explanation:
![log(x + y) = log3 + \frac{1}{2} logx+ \frac{1}{2} logy \\ \\ log(x + y) = log3 + logx ^{\frac{1}{2}} + logy ^{\frac{1}{2}}\\ \\ log(x + y) = log3 + log(xy) ^{\frac{1}{2}} \\ \\ log(x + y) = log[3(xy) ^{\frac{1}{2}}] \\ \\ x + y = 3(xy) ^{\frac{1}{2}} \\ \\ squaring \: both \: sides \\ {(x + y)}^{2} = \bigg(3(xy) ^{\frac{1}{2}} \bigg)^{2} \\ \\ {x}^{2} + {y}^{2} + 2xy = 9xy \\ \\ {x}^{2} + {y}^{2} = 9xy - 2xy \\ \\ \purple{ \bold{{x}^{2} + {y}^{2} = 7xy}} \\ thus \: proved](https://tex.z-dn.net/?f=log%28x%20%2B%20y%29%20%3D%20log3%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20logx%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20logy%20%5C%5C%20%20%5C%5C%20log%28x%20%2B%20y%29%20%3D%20log3%20%2B%20%20%20%20logx%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%2B%20%20%20logy%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%20%20%5C%5C%20%20log%28x%20%2B%20y%29%20%3D%20log3%20%2B%20%20%20%20log%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5C%5C%20%20%5C%5C%20log%28x%20%2B%20y%29%20%3D%20%20log%5B3%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5D%20%5C%5C%20%20%5C%5C%20x%20%2B%20y%20%3D%203%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5C%5C%20%20%5C%5C%20squaring%20%5C%3A%20both%20%5C%3A%20sides%20%5C%5C%20%20%7B%28x%20%2B%20y%29%7D%5E%7B2%7D%20%20%3D%20%20%5Cbigg%283%28xy%29%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cbigg%29%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%20%2B%202xy%20%3D%209xy%20%5C%5C%20%20%5C%5C%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%20%3D%209xy%20-%202xy%20%5C%5C%20%20%5C%5C%20%20%20%5Cpurple%7B%20%5Cbold%7B%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%20%3D%207xy%7D%7D%20%5C%5C%20thus%20%5C%3A%20proved)
Answer:
95% Confidence interval: (0.8449,0.9951)
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 50
Number of times the dog is right, x = 46
95% Confidence interval:
Putting the values, we get:
(0.8449,0.9951) is the required 95% confidence interval for the proportion of times the dog will be correct.