Considering the dot plot, we have that only 7 of the 25 measures are ratings of at most 6, that is, 6 is less than both the mean and the median of the distribution, hence it is not a good description of the center of this data set.
<h3>What is a dot plot?</h3>
A dot plot is a graph shows the number of times each measure appears in the data-set.
From the plot given in this exercise, it is found that only 7 of the 25 measures are ratings of at most 6, that is, 6 is less than both the mean and the median of the distribution, hence it is not a good description of the center of this data set.
More can be learned about dot plots at brainly.com/question/24912483
Answer:
Step-by-step explanation:
8.) For a triangles sides to make sense, you must be able to add up two values of the triangle, and the result should be more than the third side. Add the lowest values and see if the result is greater than the biggest number:
12.1 is less than the given side, 13.3, so a triangle cannot have the lengths.
10.) 6<x<22
To find the range for the third side of the triangle, you need to find how small x can be (the missing side) and you need to see how large it can be.
You need to see how small it can be because any two sides have to be greater than the third side. You also need to see how big it can be because, if it's too big, the other two sides will be less than the third side, which would make an open shape (see picture).
To find the range, first see how small. Subtract the known sides:
So, x has to be greater than 16.
x > 16
Now add the known sides:
x needs to be less than 28 for the other two sides to be greater than x:
x < 28
Insert the inequalities into a single inequality:
16 < x <28
X has to be greater than x, but less than 28.
Answer:
6 - (-8)
is 6 + 8
Step-by-step explanation:
1700, just use the inverse operation. The problem would look like this ( not inverse operation ) x ÷ 33 1/3 = 51
All you have to do is multiply the two numbers.
51 x 33 1/3 = 1700
Now plug it back in and check
1700 ÷ 33 1/3 does indeed equal 51.
Answer:
dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0
Step-by-step explanation:
The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out
Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.
Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.
So, dQ(t)/dt = mass flowing in - mass flowing out
dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min
Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0
So, the initial value problem is thus
dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0
