Answer:
x ≈ 52.1°
Step-by-step explanation:
Using the tangent ratio in the right triangle
tan x = =
=
, then
x = (
) ≈ 52.1° () to the nearest tenth )
What is the perimeter?
2 answers:
<h3>☆Given :</h3>
- BC = 5 in
- AC = 8in
- AB = x (in)
- Perimeter of triangle
So to find perimeter of triangle first we have to find value of x.
As it's Given one angle is 90°
So we can find value of x by using this formula:-
Hypotenuse ² = Base² + perpendicular ²
Where :-
Hypotenuse is AC
Base is BC
Perpendicular is AB
So:-
⇢AC² = BC² + AB²
⇢8² = 5² + x²
⇢64 = 5² + x²
⇢64 = 25 + x²
⇢25 + x² = 64
⇢x² = 64 - 25
⇢x² = 39 in.
⇢x = √(39)
⇢x = 6.2
Now Let's find perimeter of triangle.
We know :-
Perimeter of triangle = Side one + side two + side 3
So:-
⇢Perimeter of triangle = AB + BC + AC
⇢Perimeter of triangle = 5 + 8 + 6.2
⇢Perimeter of triangle = 13 + 6.2
⇢Perimeter of triangle = 19.2 in.
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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.
