What would the inter-rater reliability be for a 50-item measure in which the number of agreements between Rater 1 and Rater 2 was 45?
C) 0.90
Answer:
Step-by-step explanation:
Difference in the distance = 50 yards
Difference in speed = 8 - 6 = 2 y/s
<u>Time to cover the gap is:</u>
<u>Sandra will run the distance:</u>
Step-by-step explanation:
you add
Hours. Minutes
5. 48
+7. 78
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
12. 126
Then 126÷60 minutes =2hours 6mins
12hrs+2hrs=14hrs
Answer=14hours 6minutes
Given that,
Shelley's pet food store sold one customer 5 peanut butter biscuits for $3. She sold another customer 7 beef treats for $4.20.
For peanut, the cost is ratio is
For beef, the ratio is
If we take cross product of these items,
It implies, that the sale is in true proportion. From the cross product, we find that it is equal.
Outliers are data that are in a very far distance from other values in a set of data
Once an outlier is detected in a set of data, we can do the following to them:
- Discard the outlier
- Change the value of the outlier with another value within close range
- Consider the distribution given
We may have a set of data where some of the <em>values are far in distance from the majority of the data</em>. The set of such data are known as an outlier.
For example, give the set of data;
45 can be considered as an outlier since the <em>distance of data</em><em> to all other data is</em><em> large</em><em>.</em>
Once an outlier is detected in a set of data, we can do the following to them:
- Discard the outlier
- Change the value of the outlier with another value within close range
- Consider the distribution given
Learn more here: brainly.com/question/23258173