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2 years ago

7

A teacher gives a test to a large group of students. The results are closely approximated by a normal curve. The mean is 74, wi

th a standard deviation of 8. The teacher wishes to give A's to the top 8% of the students and F's to the bottom 8%. A grade of B is given to the next 16%, with D's given similarly. All other students get C's. Find the bottom cutoff for a grade of C.

1 answer:

Alinara [238K]2 years ago

6 0

Answer:

Step-by-step explanation:

Given that a teacher gives a test to a large group of students. The results are closely approximated by a normal curve

mu =74 and sigma =8

A grade starts from 100-8 = 92nd percentile

Z score for 92nd percentile = 1.405

X score = 74+8(1.405) = 85.24

--------------------

B cut off is to next 16%

Hence C would start for scores below 100-(8+16) = 76%

76th percentile = 0.705*8+74 =79.64

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For this case, what you should know is that Sue Jones insurance covers half of the expenses. Equivalently, her insurance covers:
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The question is missing parts. Here is the complete question.

Let M = $\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$ . Find $c_{1}$ and $c_{2}$ such that $M^{2}+c_{1}M+c_{2}I_{2}=0$ , where $I_{2}$ is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: $c_{1} = \frac{-16}{10}$

$c_{2}=\frac{-214}{10}$

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

$\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

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Solving equation:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

And the system of equations is:

$6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15$

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

$6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)$

$6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15$

$10c_{1} = -16$

$c_{1} = \frac{-16}{10}$

With $c_{1}$ , substitute in one of the equations and find $c_{2}$ :

$6c_{1}+c_{2}=-31$

$c_{2}=-31-6(\frac{-16}{10} )$

$c_{2}=-31+(\frac{96}{10} )$

$c_{2}=\frac{-310+96}{10}$

$c_{2}=\frac{-214}{10}$

<u>For the equation, </u> $c_{1} = \frac{-16}{10}$ <u> and </u> $c_{2}=\frac{-214}{10}$ <u />

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