Write each phrase as an algebraic expression. 16 seconds faster than Aya’s time.
1 answer:
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We are given a function f ( x ) defined as follows:
We are to determine the value of f ( x ) when,
In such cases, we plug in/substitue the given value of x into the expressed function f ( x ) as follows:
We will apply the power on both numerator and denominator as follows:
Now we evaluate ( 2 ) raised to the power of ( 1 / 9 ).
Next apply the division operation as follows:
Once, we have evaluated the answer in decimal form ( 5 decimal places ). We will round off the answer to nearest thousandths.
Rounding off to nearest thousandth means we consider the thousandth decimal place ( 3rd ). Then we have the choice of either truncating the decimal places ( 4th and onwards ). The truncation only occurs when (4th decimal place) is < 5.
However, since the (4th decimal place) = 8 > 5. Then we add ( 1 ) to the 3rd decimal place and truncate the rest of the decimal places i.e ( 4th and onwards ).
The answer to f ( 1 / 2 ) to the nearest thousandth would be:
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