When rounding 243.875: to the nearest tenth, it becomes 243.9; to the nearest hundredth, it is 243.88; to the nearest ten, it rounds down to 240; and to the nearest hundred, it rounds down to 200. The general rule in rounding states that if the decimal is less than 5, the number remains the same; if it is 5 or more, you round up.
Answer:
Attached is the histogram illustrating the marathon runners’ times.
Step-by-step explanation:
The provided data is as follows;
2.21
2.25
2.76
3.1
3.3
3.5
3.6
3.77
3.8
4.23
4.25
4.25
4.6
4.9
From this data, we can determine;
The count of runners finishing between 0 and 1 hour = 0
The count of runners finishing between 1 and 2 hours = 0
The count of runners finishing between 2 and 3 hours = 3
The count of runners finishing between 3 and 4 hours = 6
The count of runners finishing between 4 and 5 hours = 5
Based on these frequencies across the various time ranges, the histogram for the provided data has been constructed and is attached.
Answer:
At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.
Step-by-step explanation:
A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.
The null and alternative hypotheses are:

The significance level is set at 0.10.
The sample mean recorded is 17, and the sample standard deviation is 5.37.
The degrees of freedom are calculated as df=(20-1)=19.
The t-statistic is:

The two-tailed P-value corresponding to t=1.67 is P=0.11132.
<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.
At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.
Max's initial distance from Kim
Max's consistent velocity
Answer:
The correct answer is;
A. 0.17
Step-by-step explanation:
Here are the provided details;
The average time taken for a cashier to handle an order, μ = 276 seconds
The deviation from the average, σ = 38 seconds
The z-score for an order processing time of x = 240 seconds can be calculated as follows;

Thus;

The resulting probability
P(z = -0.9474) = 0.17361
Hence, the estimated proportion of orders processed in under 240 seconds is roughly 0.17361 or 0.17 when rounded to two decimal places.