The image of the histogram distribution is missing, so i have attached it.
Answer:
Option C: The distribution is unimodal. Its mode is near 60 inches
Step-by-step explanation:
Mode simply means the most occurring value in a set of data.
Now, for histogram, the mode is the highest peak represented in the bars.
Now, from the image attached, it is seen that there is a clear highest peak. Thus, the histogram has only one peak and it can be described as unimodal.
We are not given the mode height from the histogram but we can see that it is between 60 and 68 inches. And if carefully observed it will have a value closer to 60.
Thus,option C is correct.
Answer:
m<RPS = 53 degrees
Step-by-step explanation:
Find the diagram attached
From the diagram shown;
m<QPR + m<RPS = 90
Given that;
m<QPR = x
m<RPS = x+16
Substitute into the formula;
m<QPR + m<RPS = 90
x+x+16 = 90
2x+16 = 90
2x = 90-16
2x = 74
x = 74/2
x = 37degrees
Get m<RPS;
Since m<RPS = x+16
m<RPS = 37 + 16
m<RPS = 53 degrees
Answer:
16,460 gallons
Step-by-step explanation:
This is a differential equation problem, we have a constant flow of contaminant into the lake, but also we know that only a fraction of that quantity of contaminant remains because of the enzymes. For that reason, the differential equation of contaminant's flow into the lake would be:
Then, we have to integrate in order to find the equation for Q(t), as the quantity of contaminant in the lake, in function of time.
Now, we use the given conditions to replace them in the equation, in order to solve for 
Then, we reorganize the equation and we replace t for 17 hours, in order to determine the quantity of contaminant at that time:
Answer:
100°
Step-by-step explanation:
the angles are vertical, so they are equal.
-5 ÷w=?
Answer the rest on your own but I just set it up for you.
