Answer:
15.95 feets
Step-by-step explanation:
Given that:
Angle of elevation = 62°
Using trigonometric relation :
Tanθ = opposite / Adjacent
From the drawing attached :
Opposite = 30 feets
Distance between girl and lighthouse = d
Tan 62 = 30 / d
1.8807264 * d = 30
d = 30 / 1.8807264
d = 15.951283
d = 15.95 feets
Answer:
a
The 95% confidence interval is
Generally the interval above can interpreted as
There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval
b
Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable
c
Step-by-step explanation:
From the question we are told that
The sample size is n = 60
The number of stores that had more than 2 items price incorrectly is k = 52
Generally the sample proportion is mathematically represented as
=>
=>
From the question we are told the confidence level is 95% , hence the level of significance is
=>
Generally from the normal distribution table the critical value of is
Generally the margin of error is mathematically represented as
=>
=>
Generally 95% confidence interval is mathematically represented as
=>
=>
Generally the interval above can interpreted as
There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval
Considering question b
Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable
Considering question c
From the question we are told that
The margin of error is E = 0.05
From the question we are told the confidence level is 95% , hence the level of significance is
=>
Generally from the normal distribution table the critical value of is
Generally the sample size is mathematically represented as
=>
=>
Answer:
84.9
Step-by-step explanation:
The weighted mean is given by the sum of the products of each grade by its respective weight. If the first four grades correspond to 15% of the final grade each, and the final exam is equivalent to 40% of the final grade, Michael's final grade is:
Michael's final weighted mean is 84.9.