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GuDViN [60]
3 years ago
13

Amanda earned a score of 940 on a national achievement test that was normally distributed. The mean test score was 850 with a st

andard deviation of 100. What proportion of students had a higher score than Amanda? Use your z table. Question 6 options: 0.10 0.32 0.18 0.82
Mathematics
1 answer:
snow_lady [41]3 years ago
4 0
<span>In math notation, we've done this: z = (X - μ) / σ = (940 - 850) / 100 = 0.90 where z is the z-score X is Vivian's score (940) µ is the mean (850) σ is the standard deviation (100) As you may know, in a normal distribution it's expected that about 68% of all observations will fall within 1 standard deviation of the mean, 95% will fall within 2 standard deviations, and 99% will fall within 3 standard deviations. 
940 lie before the first standard deviation, in which 16.5% is above it
since 940 is 0.9 from the mean and 0.1 from the first standard deviation
so above it is 17.5 % = 0.175 or about 0.18 </span>
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A 2 over 3 because 2 and 3 both go in to 6 whereas the others don't.
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3 years ago
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Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.
sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

So, \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Isolating the X, we have

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right] -  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Resolving:

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]=\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X=\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]⁻¹·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]·\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a=\frac{2}{11}; b=\frac{7}{33}; c=\frac{-1}{11} and d=\frac{-3}{11}. Substituting:

X= \left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11}  \end{array}\right]·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Multiplying the matrices, we have

X=\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11}  \end{array}\right]

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3 years ago
Tony is trying to find the solution of the system 2x - 4y +12 using elimination 3x + 4y = 48 He wrote these steps to solve the p
kobusy [5.1K]

Answer: Solution: (12, 3)

Step-by-step explanation:

2x - 4y = 12

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Divide both sides by 5

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We can use the value of x to find y

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3 (12) + 4y = 48

36 + 4y = 48

Subtract 36 from both sides

4y = 12

Divide both sides by 4

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Solution: (12, 3)

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Irina18 [472]

Step-by-step explanation:

Hey there!

By looking at the figure, the given "2x" and "6x + 28" are co-interior angle.

So,

2x + 6x + 28 = 180°. ( sum of co-interior angle is 180°)

8x + 28°=180°

8x = 180° - 28°

8x = 152°

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X = 19°

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3 years ago
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Are the two triangles congruent?
Katena32 [7]

Answer:

yes.......................

Step-by-step explanation:

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