Answer:
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Step-by-step explanation:
Sample size, n = 51
p = 0.62
1 - p = 1 - 0.62 = 0.38
n = 515
Confidence level = 90% = Zcritical at 90% = 1.645
Confidence interval = (p ± margin of error)
Margin of Error = Zcritical * sqrt[(p(1-p))/n]
Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]
Margin of Error = 1.645 * 0.0214
Margin of Error = 0.035203
Lower boundary = (0.62 - 0.035203) = 0.584797
Upper boundary = (0.62 + 0.035203) = 0.655203
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Answer:
my answer would be D. The stock closed lower than the opening price.
Answer:
independent: day number; dependent: hours of daylight
d(t) = 12.133 +2.883sin(2π(t-80)/365.25)
1.79 fewer hours on Feb 10
Step-by-step explanation:
a) The independent variable is the day number of the year (t), and the dependent variable is daylight hours (d).
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b) The average value of the sinusoidal function for daylight hours is given as 12 hours, 8 minutes, about 12.133 hours. The amplitude of the function is given as 2 hours 53 minutes, about 2.883 hours. Without too much error, we can assume the year length is 365.25 days, so that is the period of the function,
March 21 is day 80 of the year, so that will be the horizontal offset of the function. Putting these values into the form ...
d(t) = (average value) +(amplitude)sin(2π/(period)·(t -offset days))
d(t) = 12.133 +2.883sin(2π(t-80)/365.25)
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c) d(41) = 10.34, so February 10 will have ...
12.13 -10.34 = 1.79
hours less daylight.
Answer:
Alright well simplify the expression which gives you
Exact form: 1495/32
Mixed number form: 46 23/32 Hope this help's :)
Step-by-step explanation:
Answer/Step-by-step explanation:
a. The the class width for the 1st, 2nd, 3rd, and 5th row equal 5 in the frequency distribution. Whereas, in the 4th row, the class width is 4 (49 - 45). This makes the construction incorrect.
b. Here, the upper limit of each consecutive class was used as the lower limit if the next class. For example, 9 is a member of the first class, which is the upper limit if the class in the first row, it ought not to be included as a member of the next class in the second row.
c. In the 3rd row, the lower limit of the class should be 133, NOT 138.
d. Different class width were used.
Row 1 class width = 13 - 9 = 4
Row 2 class width = 19 - 14 = 5
Row 3 class width = 25 - 20 = 5
Row 4 class width = 28 - 26 = 2
Row 5 class width = 32 - 29 = 3
