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KatRina [158]
3 years ago
13

In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co

llege, and 30% of the students were in the Education College. To see whether the proportions have changed, a random sample of 300 students from ABC University was selected. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. Refer to Exhibit 12-4. This problem is an example of a _____. a. multinomial population b. normally distributed variable c. uniformly distributed variable
Mathematics
1 answer:
KIM [24]3 years ago
4 0

Answer:

A. Multinomial population

Explanation:

A multinomial population has categorical data. In other words data in this form has different classes in which data is grouped into. These classes do not overlap and are more than two as opposed to binomial populations that have just two outcome variables or data categories. For example in the above data, we see that data is grouped into three categories:students in business college, students in liberal art college and students in education college

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Darya [45]

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2 \tan (6x)+2 \sec (6x)+\text{C}

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

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If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x

\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}

If the terms are multiplied by constants, take them outside the integral:

\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x

Multiply by the conjugate of 1 - sin(6x) :

\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x

\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x

\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:

\implies \sin^2 (6x) + \cos^2 (6x)=1

\implies \cos^2 (6x)=1- \sin^2 (6x)

\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x

Expand:

\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x

\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:

\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x

\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x

\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}

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\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}

Simplify:

\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}

\implies 2 \tan (6x)+2 \sec (6x)+\text{C}

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

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