Slope is 2/3
intercept is (0,3)
y=2/3 x + 3
It's C because the equations are similar
Answer:
Option E is correct.
The expected number of meals expected to be served on Wednesday in week 5 = 74.2
Step-by-step Explanation:
We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.
Week
Day 1 2 3 4 | Total
Sunday 40 35 39 43 | 157
Monday 54 55 51 59 | 219
Tuesday 61 60 65 64 | 250
Wednesday 72 77 78 69 | 296
Thursday 89 80 81 79 | 329
Friday 91 90 99 95 | 375
Saturday 80 82 81 83 | 326
Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952
Total number of meals served at lunch on Wednesdays over the 4 weeks = 296
Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443
Total number of meals expected to be served in week 5 = 490
Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3
Checking the options,
74.3 ≈ 74.2
Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2
Hope this Helps!!!
Answer:
35, 43
Step-by-step explanation:
The pattern is that to obtain a term in the sequence , add 8 to the previous term, then
27 + 8 = 35
35 + 8 = 43
35 and 43 are the next 2 terms in the sequence
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.
