NEED HELP ASAP WILL MARK BRAINLIEST
1 answer:
You might be interested in
Answer:
Part A)
The height of the water level in the rectangular vessel is 2 centimeters.
Part B)
4000 cubic centimeters or 4 liters of water.
Step-by-step explanation:
We are given a cubical vessel that has side lengths of 10cm. The vessel is completely filled with water.
Therefore, the total volume of water in the cubical vessel is:
This volume is poured into a rectangular vessel that has a length of 25cm, breadth of 20cm, and a height of 10cm.
Therefore, if the water level is h centimeters, then the volume of the rectangular vessel is:
Since the cubical vessel has 1000 cubic centimeters of water, this means that when we pour the water from the cubical vessel into the rectangular vessel, the volume of the rectangular vessel will also be 1000 cubic centimeters. Hence:
Therefore:
So, the height of the water level in the rectangular vessel is 2 centimeters.
To find how how much more water is needed to completely fill the rectangular vessel, we can find the maximum volume of the rectangular vessel and then subtract the volume already in there (1000 cubic centimeters) from the maximum volume.
The maximum value of the rectangular vessel is given by :
Since we already have 1000 cubic centimeters of water in the vessel, this means that in order to fill the rectangular vessel, we will need an additional:
Sincer 1000 cubic centimeters is 1 liter, this means that we will need four more liters of water in order to fill the rectangular vessel.
Answer:
The complete question in the graphic
A.
After synchronized exercise, it was observed that the average proximity rate has increased, which shows that the mean of the variable is positive and there is high synchronization in the groups.
.
B.
If groups with low synchronization and high effort are compared, the result of the average will be positive, so the average proximity rate will also increase after the synchronization exercise is performed.
On the contrary, if we observe the groups that have low synchronization and low effort, the mean of the variable is negative, thus the average proximity rate after performing the exercise will decrease.
(C)
considering that the total degrees of freedom is 259, it will show 260 students in the analysis.
D.
With a significance level of 5% or 0.05, the P value is less than the significance level, which leads to differentiating the mean of the scores before and after performing the combinations.
E.
With a significance level of 1% or 0.01, the p-value is higher than the significance level. With this data we can conclude that there is no difference between the results before and after the synchronized exercise
