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Feliz [49]
3 years ago
13

A college professor reports that students who finish exams early tend to get better grades than students who hold on to exams un

til the last possible moment. The correlation between exam score and amount of time spent on the exam is an example of a ____.​
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:

The correlation between exam score and amount of time spent on the exam is an example of a <u>negative correlation</u>.​

Step-by-step explanation:

Consider the provided statement.

College professor reports that students who finish exams early tend to get better grades than students who hold on to exams until the last possible moment.

That means the student who finish the exam early will get the highest marks. One who finish after the first student will get second highest mark and the one who finish in the end will get the least marks.

Now consider time as first variable and marks as second variable.

That means as the first variable increase the second variable decreases.

According to the definition of Negative correlation: It is a relationship between two variables in which one variable increases as the other decreases, and vice versa.

Hence, the correct answer is negative correlation.

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A rumor spreads through a small town. Let y ( t ) be the fraction of the population that has heard the rumor at time t and assum
Ivan

Answer:

Differential equation

\frac{dy}{dt} =ky(1-y)

Solution

y=\frac{1}{1+4e^{-0.327t}}

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

\frac{dy}{dt} =ky(1-y)

Solving the differential equation

\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}

Initial conditions:

y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48

The rumor reaches 80% at 8.48 days.

8 0
3 years ago
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