This question is not complete because is lacks the options for each question.
Complete question
mr. whittaker’s science class uses tide gauges to measure annual variations in water levels at different parts of a river, and then compares those variations to the average annual trend. matea recorded that the water level in one part of the river fell 1.05 millimeters per year for 2.48 years. this data will be compared to the average annual trend, which shows the water level rising 1.8 mm/year.
1) Which number represents the rate at which the water level fell?
a. -1.05 mm/year
b. -1.8 mm/year
c. -2.48 years
d. -2.48 mm/year
2) Which number should the rate be multiplied by to find the total variation in water level?
a. 1.05 mm/year
b. 1.8 mm/year
c. 2.48 years
d. 2.48 mm/year
Answer:
1) Which number represents the rate at which the water level fell?
a. -1.05 mm/year
2) Which number should the rate be multiplied by to find the total variation in water level?
c. 2.48 years
Step-by-step explanation:
Total Variation in the water level is calculated by multiplying the rate at which the water level and number or years at which the rate of the water level fell
From the question, we can see that the rate at which the water level fell is -1.05 mm/year and number of years at which the water level fell is 2.48mm
Hence,Total Variation in the water level is calculated as
-1.05 mm/year × 2.48 years
Total variation in the water level = -2.604mm
Answer:
0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x
Step-by-step explanation:
Given the data in the question;
sample size n = 28
slope of the least squares regression line of y on x or sample estimate = 0.0623
standard error = 0.0224
95% confidence interval
level of significance ∝ = 1 - 95% = 1 - 0.95 = 0.05
degree of freedom df = n - 2 = 28 - 2 = 26
∴ the equation will be;
⇒ sample estimate ± ( t-test) ( standard error )
⇒ sample estimate ± ( 
) ( standard error )
⇒ sample estimate ± ( 
) ( standard error )
⇒ sample estimate ± ( 
) ( standard error )
{ from t table; ( ) = 2.055529 = 2.056
so we substitute
⇒ 0.0623 ± ( 2.056 )( 0.0224 )
Therefore, 0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x
Tens refers to the unit 10,
This means that 45 10s is equal to:
45*10 = 450
Hope this helps! :)
Not the best, teachers are so annoying lol. Hope you are doing well though, don’t give up❤️
