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siniylev [52]
4 years ago
14

Suppose that you are asked to conduct a study to determine whether smaller class

Mathematics
1 answer:
weeeeeb [17]4 years ago
8 0

Answer:

a. correlation

b. inverse linear correlation exists If the higher the population of students lead to a decrease in test score,

c. yes

Step-by-step explanation:

a. Correlation is a measure of the amount of association existing between two variables.

b. For linear correlation, if points are plotted on a graph and all the points lie on a straight line, then perfect linear correlation is said to exist. When a straight line having a positive gradient can reasonably be drawn through points on a graph positive or direct linear correlation exists,

Similarly,when a straight line having a negative gradient can reasonably be drawn through points on a graph, negative or inverse linear correlation exists,

The results of this determination give values of r lying between +1 and −1, where +1 indicates perfect direct  Positive linear correlation and −1 indicates perfect inverse correlation or Negative linear correlation and 0 indicates that no correlation exists.

If the higher the population of students lead to a decrease in test score, there will definitely be a negative correlation between class size and test score. i.e low class size result in high test score which consequently lead to high performance.

c. YES

A negative correlation means low class size result in high test score which consequently lead to better performance.

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8 0
3 years ago
Three fifths of what number equals one
Feliz [49]
21 because it is being multiplied
6 0
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mihalych1998 [28]

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7 0
4 years ago
WILL GIVE BRAINLIEST! HELP ASAP!
makkiz [27]

Answer:

any nuber less then 3.5  and 5th awnser

Step-by-step explanation:

x is less then 3.5 so any    number less then 3.5

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7 0
3 years ago
 The distance between city A and B is 600 km. The first train left A and headed towards B at the speed of 60 km/hour. The secon
hoa [83]
\bf \begin{array}{ccccllll}
&distance&rate(km/hr)&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
A&d&60&t\\
B&600-d&v&t+3
\end{array}

\bf \textit{meaning}\implies 
\begin{cases}
d=(60)(t)
\\ \quad \\
600-d=(v)(t+3)\\
------------\\
d=\boxed{60t}\qquad thus
\\ \quad \\
600-\boxed{60t}=v(t+3)\leftarrow \textit{solve for "t"}
\end{cases}

keep in mind, that "t" is the time when the train at A station, left towards B station

they met, at some time "t", and by the time that happened, train from A
which started 3 hours earlier, had already covered "d" distance,
whatever that is
and the train coming from B, covered, 600-d, or the difference
8 0
3 years ago
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