Answer:
Mr. Rodriguez made a reasonable inference. His sampling was random, nonbiased, and reasonably large compared to the school population. 12 is 20% of 60.
Step-by-step explanation:
the ratio in the sample is 12/60, or 0.2
Answer:
The fair assumption was made by Mr. Rodriguez. His surveys were random, non-biased and very large according to the population of the schools. Of 60, 12 is 20%.
Im sad
B. 27x1/27x2
C. 3x3x3/3x3x3x2
D. 1/2
Answer: C. 27/54 .
Step-by-step explanation:
Here, given number, 27/54
Since the prime factor form of 27 =
And, the prime factor form of 54 =
putting the prime factor form of both 27 and 54 in the given numbers.
we get,
Thus, by putting the prime factor form by C we get the simplest form of the given fraction.
Therefore, Option C is correct.
I apologize for the oversight. Let's solve it and provide the answer.
Step-by-step explanation :
1. **Standard Error Calculation :
\[ SE = \frac{\sigma}{\sqrt{n}} \]
\[ SE = \frac{25.3}{\sqrt{215}} \]
\[ SE \approx 1.7292 \]
2. **Z-Score Calculation:**
\[ z = \frac{X - \mu}{SE} \]
\[ z = \frac{197.2 - 195.5}{1.7292} \]
\[ z \approx 0.9836 \]
3. Finding the Probability :
To find the probability that the sample mean is greater than 197.2, we need to find the area to the right of the z-score in the standard normal distribution table.
For \( z \approx 0.9836 \), the area to the left is approximately \( 0.8374 \).
Since we want the area to the right (the probability the sample mean is greater than 197.2), we need to subtract that value from 1 :
\[ P(X > 197.2) = 1 - 0.8374 \]
\[ P(X > 197.2) = 0.1626 \]
Answer : The probability that the sample mean is greater than 197.2 is approximately \( 0.1626 \) or \( 16.26\% \).