LOTS OF POINTS HELP

Mathematics

2 answers:

Lesechka [4]3 years ago

8 0

Answer:

Jessica has 48 different sets of classes to choose from.

Step-by-step explanation:

Given

The four categories are:

Economics: 5

Mathematics: 4

Education: 3

Sociology: 2

Suppose that sections for the 3 most popular classes in Economics are full.

Economics is left with (5-3) choices

Economics = 2 choices

Since there are equal chances of making selections, we multiply the number of options in each category to get the total set of classes Jessica can chose from.

Number of choices = 2 * 4 * 3 * 2

Number of choices = 48 choices

Hence, Jessica has 48 different sets of classes to choose from.

Volgvan3 years ago

5 0

<span>Category   Number of Choices
Economics   5 - 3 = 2
Mathematics 4
Education 3
Sociology   2

Since out of 5 sections in Economics 3 are full, there are only 2 remaining sections available for Jessica to choose from.

I think this is a combination:

n = 2 + 4 + 3 + 2 = 11
r = 1 + 1 + 1 + 1 = 4

nCr = n! / r!(n-r)! = 11! / 4!(11-4)! = 11! / 4! * 7! = 330

Jessica can have 330 different sets of classes to take.

</span>

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If there are 3 birds and one get shot at how many how left(trick question)

sasho [114]

Answer:

3 birds

Step-by-step explanation:

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8 0

2 years ago

What is the period and midline?

OverLord2011 [107]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}\qquad$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)