The correct answer is:
2u-v=7(5i-3j)
Therefore, 5, and 3 go inside the box.
Aruthmetic sequene is
an=a1+(n-1)d
where d=common difference between terms
adds 6 every time
d=6
first term is 8
a1=8
8+6(n-1)
distribute
8+6n-6
8-6+6n
2+6n is answer
Answer:
since -3.73 is less than 1.645, we reject H₀.
Therefore this indicate that the proposed warranty should be modified
Step-by-step explanation:
Given that the data in the question;
p" = 13/20 = 0.65
Now the test hypothesis;
H₀ : p = 0.9
Hₐ : p < 0.9
Now lets determine the test statistic;
Z = (p" - p ) / √[p×(1-p)/n]
= (0.65 - 0.9) /√[0.9 × (1 - 0.9) / 20]
= -0.25 / √[0.9 × 0.1 / 20 ]
= -0.25 / √0.0045
= -0.25 / 0.067
= - 3.73
Now given that a = 0.05,
the critical value is Z(0.05) = 1.645 (form standard normal table)
Now since -3.73 is less than 1.645, we reject H₀.
Therefore this indicate that the proposed warranty should be modified
F(x) = (x - 4) (x^2 + 4) would be your answer.
Answer:
m=rise over run
so it would be 
Step-by-step explanation:
