Just took the quiz.
A. 145°
C. /
A. 250°
D. -.87,-0.5
A. Sqrt(3)/2
Answer:
A normal distribution or z-test is used to construct a confidence interval.
Step-by-step explanation:
We are given the following in the question:
Sample mean,
= $3120
Sample size, n = 40
Population standard deviation, σ = $677
The distribution of earnings of college is a normal distribution.
Conditions:
- Since we are given the population standard deviation and the the sample size is also greater than 30.
Conclusion:
Thus, we use a normal distribution or z-test to construct a confidence interval.
Answer:
28 hours
Step-by-step explanation:
let the time period at which the temperature be equal in both Brownsville and Mesquite Texas be 'x'.
Given,
In Brownsville Texas, the temperature is expected to drop 1.5° each hour and In Mesquite Texas, the temperature is expected to drop 2° each hour.
Now, according to the question,
after 'x' hours,
⇒ 80° - 1.5°(x) = 94° - 2°(x)
⇒ 0.5°(x) = 14°
⇒ x = 28
Hence, Till 28 more hours the temperature in Mesquite Texas be greater than that in Brownsville.