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2 years ago

10

Last year your professor wanted to study the pattern in class grades. The following table contains a sample of grades collected

from the final exam last semester:
64 80 75 98 75

Using Table and looking at the relative positions of the Mean, Median and Mode for the sample, what can you say about the distribution?

1 answer:

Maslowich2 years ago

8 0

Answer:

mean = 78.4

median = 77.5

mode = 75

This is Right - skewed (positive skewness) distribution

Step-by-step explanation:

<u>Mean:-</u>

The mean (average) is found by adding all of the numbers together and dividing by the number of items and it is denoted by x⁻

mean = $\frac{64+80+75+98+75}{5}$

mean (x⁻ ) = 78.4

The mean of the given data = 78.4

<u>Median:</u>

The median is found by ordering the set from lowest to highest and finding the exact middle.

64 ,75, 80, 98

The middle term of the given data set = $\frac{75+80}{2} =77.5$

<u>Mode :</u>

The mode is the most common repeated number in a data set.

64 ,75, 75, 80, 98

in data the most common number = 75

<u>Conclusion</u>:-

mean = 78.4

median = 77.5

mode = 75

This is Right - skewed (positive skewness) distribution

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Step-by-step explanation:

One way to answer this question is to use synthetic division to find the remainder from division of the polynomial by (x-3). If the polynomial is written in Horner form, evaluating the polynomial for x=3 is substantially similar.

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2 years ago

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Answer:

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Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

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$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

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We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

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Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

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Writing the matrix A :

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Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

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$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

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