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weqwewe [10]
2 years ago
14

What type of association does the scatterplot appear to have?

Mathematics
2 answers:
inn [45]2 years ago
8 0

Answer:

positive linear association

Step-by-step explanation:

Sliva [168]2 years ago
8 0

Answer:

A scatter plot shows the association between two variables. A scatter plot matrix shows all pairwise scatter plots for many variables.

Step-by-step explanation:

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There is a set of 100 observations with a mean of 50 and a standard deviation of 0. What is the value of the smallest observatio
neonofarm [45]
In order to obtain a standard deviation of 0, all members of the data set should be equal.
This means that each of the 100 members in the data set is equal to 50 (the mean value).

Answer:
The smallest observation in the set is 50.
8 0
2 years ago
Read 2 more answers
Determined to find the slope<br>(1 7K), and (-3,5K)​
Helga [31]

Answer: \dfrac{K}{2} .

Step-by-step explanation:

As we know ,

The slope of a line that passes through (x_1,y_1) and (x_2,y_2)  is given by :

\dfrac{y_2-y_1}{x_2-x_1}

The slope of a line that passes through (1 , 7K), and (-3,5K)​ =

\dfrac{5K-7K}{-3-1}\\\\=\dfrac{-2K}{-4}\\\\=\dfrac{K}{2}

Hence, the slope of the given line is \dfrac{K}{2} .

8 0
3 years ago
The yearbook club had a meeting. The meeting had 20 people, which is four-fifths of the club. How many people are in the club?
Rama09 [41]
There are 32 people in the yearbook club

4 0
2 years ago
Read 2 more answers
Through the point (3,1.2) that has an y intercept of 3
topjm [15]

Answer:

y = -0.6x +3

Step-by-step explanation:

equation of the line through point (3, 1.2) and y-interecept of 3

y= mx+b

m is the slope and b is thhe y-intercept

y = mx + 3

between points (3, 1.2) and (0,3) the slope is

m= (y2-y1)/ (x2-x1) = (1.2-3) / (3-0) = -1.8/3 = -0.6

y = -0.6x +3

3 0
2 years ago
I don’t know how to do this
stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as x approaches the edges. For example,

g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}

has two pieces, 2x+5 and x^2-10, both of which are continuous by themselves on the provided intervals. In order for g to be continuous everywhere, we need to have

\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)

By definition of g, we have g(-3)=2(-3)+5=-1, and the limits are

\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1

\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1

The limits match, so g is continuous.

For the others: Each of the individual pieces of f,h are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.

4 0
3 years ago
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