Answer:
Step-by-step explanation:
(a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3cb² +3 ac² + 3bc² + 6abc
a = 5a ; b =y ; c = z
(5x + y + z)(5z + y + z )(5z + y +z) = (5x + y +z)³
= (5x)³ + y³ +z³ + 3(5x)²y + 3(5x)²z + 3(5x)*y² + 3*z*y² + 3*5x*z² + 3*y*z² + 6*5x*y*z
= 125x³ + y³ +z³ + 3*25x²y + 3*25x²*z + 15xy² + 3zy² + 15xz² + 3yz² + 35xyz
= 125x³ + y³ + z³ + 75x²y + 75x²z + 15xy² + 3zy² + 15xz² + 3yz² + 35xyz
Answer:
for 18 x is 6 and for 20 it is 11
Step-by-step explanation:
For 18, since -1+2x and 5+x is the same thing, set up and equation. WE can simplify to -1+x=5 and then we get x=6
For 20, 18 is 11 since 18=-4+2x. If we simplify we get 22=2x then we can use algebra to say that x=11
Hope this helps :)
Where’s the question I need to see the problem
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:
the slope is 0
Step-by-step explanation: