Answer:


Step-by-step explanation:
One is given the following function:

One is asked to evaluate the function for
, substitute
in place of
, and simplify to evaluate:



A recursive formula is another method used to represent the formula of a sequence such that each term is expressed as a function of the last term in the sequence. In this case, one is asked to find the recursive formula of an arithmetic sequence: that is, a sequence of numbers where the difference between any two consecutive terms is constant. The following general formula is used to represent the recursive formula of an arithmetic sequence:

Where (
) is the evaluator term (
) represents the term before the evaluator term, and (d) represents the common difference (the result attained from subtracting two consecutive terms). In this case (and in the case for most arithmetic sequences), the common difference can be found in the standard formula of the function. It is the coefficient of the variable (n) or the input variable. Substitute this into the recursive formula, then rewrite the recursive formula such that it suits the needs of the given problem,



Answer:
2
Step-by-step explanation:
Divide 3.4 by 1.7 and you get 2
No! 11, the input, has two outputs which makes it not a function! :)
2w+2(5+2w)=33
2w+10+4w=33
6w=22
w=11/3 or 3 and 2/3
Answer:
We <em>fail to reject H₀ </em>as there is insufficient evidence at 0.5% level of significance to conclude that the mean hours of TV watched per day differs from the claim.
Step-by-step explanation:
This is a two-tailed test.
We first need to calculate the test statistic. The test statistic is calculated as follows:
Z_calc = X - μ₀ / (s /√n)
where
- X is the mean number of hours
- μ₀ is the mean that the sociologist claims is true
- s is the standard deviation
- n is the sample size
Therefore,
Z_calc = (3.02 - 3) / (2.64 /√(1326))
= 0.2759
Now we have to calculate the z-value. The z-value is calculated as follows:
z_α/2 = z_(0.05/2) = z_0.025
Using the p-value method:
P = 1 - α/2
= 1 - 0.025
= 0.975
Thus, using the positive z-table, you will find that the z-value is
1.96.
Therefore, we reject H₀ if | Z_calc | > z_(α/2)
Thus, since
| Z_calc | < 1.96, we <em>fail to reject H₀ </em>as there is insufficient evidence at 0.5% level of significance to conclude that the mean hours of TV watched per day differs from the claim.