There was 3 erasers in each pack because 2 times 3 is 6 plus 2 equals 8. so there are 3 in each pack.
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
Answer:
Step-by-step explanation:
Your equation is:
You also know that x = -3 and y = 2. You can rewrite the equation by substituting x and y with their actual values:
First, you want to do the exponentiation (meaning you want to calculate and parts). equals 9 and equals 16.
So the equation now is:
Now you want to do multiplication. equals 36 and equals 32. So you're left with:
Which equals 68.
Answer:
Jane is 52 7/40 inches tall
Step-by-step explanation:
What you do is you subtract 1 3/8 from 54 3/4 which gives you Carl's height, which is 53 3/8.
Lastly, you subtract 1 1/5 from 53 3/8 which is 52 7/40.
Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1
Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;
so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
;