By using trigonometric relations, we will see that x = 9.97°.
<h3>
How to find the missing angle?</h3>
First, we need to find the bottom cathetus of the smaller triangle, we will use the relation:
Tan(θ) = (opposite cathetus)/(adjacent cathetus).
Where:
- θ = 26°
- Adjacent cathetus = k
- Opposite cathetus  = 55ft.
Replacing that we get:
Tan(26°) = 50ft/k
Solving this for k, we get:
k = 55ft/tan(26°) = 112.8 ft
Now, we can see that the longer triangle adds 200ft to this cathetus, so now we will have:
- angle = x
- opposite cathetus = 55ft
- adjacent cathetus = 112.8ft + 200ft = 312.8ft.
Then we have:
Tan(x) = (55ft/312.8ft)
Using the inverse tangent function in both sides, we get:
x = Atan(55ft/312.8ft) = 9.97°
If you want to learn more about right triangles, you can read:
brainly.com/question/2217700
 
        
             
        
        
        
Answer:
how many square centimeters is the box??
Step-by-step explanation:
 
        
             
        
        
        
There is no rectangle shown, nor ruler given to measure and attempt your work. Nevertheless, I will illustrate how to find the volume of a rectangular prism.
Step-by-step explanation:
Given a rectangular prism of sides
Length = l 
Width = w
Height = h
The volume of this prism is given as
V = l × b × h. 
Example, if 
Length = 4cm
Width = 3cm
Height = 2cm
Volume = 4 × 3 × 2 
= 24cm³.
 
        
             
        
        
        
Answer:
<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938
</em>
Step-by-step explanation:
<em>Given:</em>
<em>Here Section 1 students = 20
</em>
<em>
Section 2 students = 30
</em>
<em>
Here there are 15 graded exam papers.
</em>
<em>
(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070
</em>
<em>
(b) Here if x is the number of students copies of section 2 out of 15 exam papers.
</em>
<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15
</em>
<em>Then,
</em>
<em>
Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798
</em>
<em>
(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)
</em>
<em>
so,
</em>
<em>
Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938
</em>
<em>
Note : Here the given distribution is Hyper-geometric distribution
</em>
<em>
where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>