In detail: Based on the central limit theorem, the distribution appears normal due to the large sample size. The confidence interval is presented in the format: (Sample mean - margin of error, sample mean + margin of error). The sample mean, denoted as x, serves as the point estimate for the population mean. The confidence interval is computed as: mean ± z × σ/√n, where σ represents the population standard deviation. The formula transforms into confidence interval = x ± z × σ/√n, with specific values: x = $75, σ = $24. To find the z score, we subtract the confidence level from 100% which gives α as 1 - 0.96 = 0.04; halving this results in α/2 = 0.02, signifying the tail areas. To ensure we account for the center area, we have 1 - 0.02 = 0.98, corresponding to a z score of 2.05 for the 96% confidence level. The confidence interval becomes 75 ± 2.05 × 24/√64 = 75 ± 2.05 × 3 = 75 ± 6.15. The lower limit is 75 - 6.15 = 68.85, while the upper limit stands at 75 + 6.15 = 81.15. For n = 400, with x = $75 and σ = $24, the z score remains 2.05, resulting in the confidence interval calculated as 75 ± 2.05 × 24/√400 = 75 ± 2.05 × 1.2 = 75 ± 2.46. Subsequently, the lower bound becomes 75 - 2.46 = 72.54, and the upper limit adds up to 75 + 2.46 = 77.46. Lastly, when n = 400, x = $200, and σ = $80, the z score tied to a 94% confidence level is 1.88. Thus, the confidence interval is expressed as 200 ± 1.88 × 80/√400 = 200 ± 1.88 × 4 = 200 ± 7.52, giving us a margin of error of 7.52.
Response:
The probability that a student has a pet, given they do not have any siblings is:
Option: D ( 60%)
Step-by-step breakdown:
Let A represent the situation where a student lacks a sibling.
Let B signify the occurrence that a student has a pet.
Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.
Let P denote the chance of an event happening.
We need to determine:
P(B|A)
From our knowledge:
From the data provided:
P(A)=0.25
and P(A∩B)=0.15
Thus,
which expressed as a percentage is:
Therefore, the probability is:
60%
The missing value associated with 
is 
Explanation:
The provided equation for the table is 
This table consists of 2 columns and 5 rows.
<pconsequently it="" states="">
x y
-2 10
-1 ---
0 2
1 -2
2 -6
To find the value for y when
The value of y can be ascertained by substituting into the equation
<pthus we="" determine="">
<pmultiplying the="" term="" within="" brackets="" yields="">
<psumming the="" terms="" we="" find="">
<pso the="">value of y when
is 6.Hence, the value for is
</pso></psumming></pmultiplying></pthus></pconsequently>
