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Leto [7]
2 years ago
13

STATISTICS in a recent year the mean score on the mathematics section of the SAT test was 515 and the standard deviation was 114

. This means that people within one deviation of the mean have SAT math scores that are no more than 114 points higher or 114 points lower than the mean A. write an absolute value inequality to find the range of SAT mathematics test scores within one standard deviation of the mean B. what is the range of SAT mathematics test scores 12 standard deviation from the mean?
Mathematics
1 answer:
vesna_86 [32]2 years ago
3 0

An absolute value inequality to find the range of SAT mathematics test scores within one standard deviation of the mean is; |x – 515| ≤ 114

<h3>How to Write Inequalities?</h3>

A) We are told that;

Mean score = 515

Standard deviation = 114

We are now given that people within one deviation of the mean have SAT math scores that are no more than 114 points higher or 114 points lower than the mean. Thus, the absolute value inequality is;

|x – 515| ≤ 114

B) The range of scores to within ±2 standard deviations of the mean is;

Range = 515 ± 2(114)

Range = 287 to 743

Read more about Inequalities at; brainly.com/question/25275758

#SPJ1

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Using the determinant method, the cross product is

\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\3&8&-6\\-4&-2&-3\end{vmatrix}=\begin{vmatrix}8&-6\\-2&-3end{vmatrix}\,\vec\imath-\begin{vmatrix}3&-6\\-4&-3\end{vmatrix}\,\vec\jmath+\begin{vmatrix}3&8\\-4&-2\end{vmatrix}\,\vec k=-36\,\vec\imath+33\,\vec\jmath+26\,\vec k

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Or you can apply the properties of the cross product. By distributivity, we have

(3i + 8j - 6k) x (-4i - 2j - 3k)

= -12(i x i) - 32(j x i) + 24(k x i) - 6(i x j) - 16(j x j) + 12(k x j) - 9(i x k) - 24(j x k) + 18(k x k)

Now recall that

  • (i x i) = (j x j) = (k x k) = 0 (the zero vector)
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Putting these rules together, we get

(3i + 8j - 6k) x (-4i - 2j - 3k)

= -32(-k) + 24j - 6k + 12(-i) - 9(-j) - 24i

= (-12 - 24)i + (24 + 9)j + (32 - 6)k

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