What is the simplest form of the expression (5x + 3xy + 4y) + (4x – 2xy – 2y)? A. x + xy + 2y B. x2 + xy + 2y C. 9x + 5xy + 2y D
2 answers:
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To solve this problem you must apply the proccedure shown below:
1. You have the following expression given in the problem above:
![\sqrt[3]{216 x^{27} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%20x%5E%7B27%7D%20%7D%20)
2. Rewriting the expression we have:
![\sqrt[3]{6^3 x^{27} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B6%5E3%20x%5E%7B27%7D%20%7D%20)
3. You have that
and the exponent 
are divisible by index 
. Therefore, you have:
![\sqrt[3]{216 x^{27} } =6 x^{9}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%20x%5E%7B27%7D%20%7D%20%3D6%20x%5E%7B9%7D%20)
Therefore, as you can see,
the answer is the option, which is:
1. You have the following expression given in the problem above:
2. Rewriting the expression we have:
3. You have that
Therefore, as you can see,
the answer is the option, which is:
Answer:
The students have to score 0.74 standard deviations above the mean to be publicly recognized.
Step-by-step explanation:
A random variable <em>X</em> is said to have a normal distribution with parameters <em>µ</em> (mean) and <em>σ</em>² (variance).
If , then
, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is,
.
The distribution of these <em>z</em>-scores is known as the standard normal distribution.
The <em>z</em>-score is a standardized form of the raw score, <em>X</em>. It is a numerical measurement of the relationship between a value (<em>X</em>) and the mean (<em>µ</em>) in terms of the standard deviation (<em>σ</em>). A <em>z</em>-score of -1 implies that the data value is 1 standard deviation below the mean. And a <em>z</em>-score of 1 implies that the data value is 1 standard deviation above the mean.
Let <em>X</em> be defined as the scores of students at the National Financial Capability Challenge Exam.
It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.
That is, P (X > x) = 0.23.
⇒
⇒
The value of <em>z</em> for this probability value is:
<em>z</em> = 0.74.
*Use a <em>z</em>-table.
Thus, the students have to score 0.74 standard deviations above the mean to be publicly recognized.
