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nlexa [21]
1 year ago
9

A frequency table of grades has five classes (A, B, C, D,F) with frequencies of 4, 10, 17, 6, and respectively. Using percentage

s, what are the relativefrequencies of the five classes?Complete the table.Grade Frequency Relative frequencyA4%B 10%%D 6%2%(Round to two decimal places as needed.)с17F

Mathematics
1 answer:
fiasKO [112]1 year ago
4 0

We have the frequencies for each of the grades. We can estimate the number of students graded by adding all those frequencies. Let's call N the total number of grades:

\begin{gathered} N=4+10+17+6+2 \\ N=39 \end{gathered}

We have then a total number of grades of 39.

The corresponding relative frequency for a grade is the ratio of the frequency to the total number of "samples", 39 in this case.

Then, for grade A, the relative frequency (RF) will be:

RF_A=\frac{4}{39}\approx0.10256\approx10.26\text{ \%}

This will be the fraction of the total grades that are A. Represented as a percentage will be 10.26%, rounded to two decimal places.

Now, to complete the table we do the same for the other frequencies:

For grade B:

RF_B=\frac{10}{39}\approx0.2564\approx25.64\text{ \%}

For grade C:

RF_C=\frac{17}{39}\approx0.4359\approx43.59\text{ \%}

For grade D:

RF_D=\frac{6}{39}\approx0.1538\approx15.38\text{ \%}

For grade F:

RF_D=\frac{2}{39}\approx0.0513\approx5.13\text{ \%}

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Section 5.2 Problem 19:
rjkz [21]

Answer:

y(x)=2xe^{3x} (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation am^2+bm+c=0 where the values of m are the roots:

y''-6y'+9y=0\\\\m^2-6m+9=0\\\\(m-3)^2=0\\\\m-3=0\\\\m=3

Since the values of m are equal real roots, then the general solution is y(x)=C_1e^{m_1x}+C_2xe^{m_1x}.

Thus, the general solution for our given differential equation is y(x)=C_1e^{3x}+C_2xe^{3x}.

To account for both initial conditions, take the derivative of y(x), thus, y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}

Now, we can create our system of equations given our initial conditions:

y(x)=C_1e^{3x}+C_2xe^{3x}\\ \\y(0)=C_1e^{3(0)}+\frac{C_2}{6}(0)e^{3(0)}=0\\ \\C_1=0

y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}\\\\y'(0)=3C_1e^{3(0)}+C_2e^{3(0)}+3C_2(0)e^{3(0)}=2\\\\3C_1+C_2=2

We then solve the system of equations, which becomes easy since we already know that C_1=0:

3C_1+C_2=2\\\\3(0)+C_2=2\\\\C_2=2

Thus, our final solution is:

y(x)=C_1e^{3x}+C_2xe^{3x}\\\\y(x)=2xe^{3x}

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