Answer:
yes, the student who has 8 pets.
Step-by-step explanation:
the majority of the class has between 0-4 pets leaving the student with 8 to be the "outlier"
Answer:
0.800
Step-by-step explanation:
Hey there!
![\left[\begin{array}{ccc}\boxed{\boxed{ \frac{6}{4}}} \end{array}\right] = \left[\begin{array}{ccc}\boxed{\boxed{150}}\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B%20%5Cfrac%7B6%7D%7B4%7D%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B150%7D%7D%5Cend%7Barray%7D%5Cright%5D%20)
All I did was to multiply each side by 100 which then go me this!
Hope this helps!
Sin51=y/12
y=12sin51 units
y≈9.33 units (to nearest hundredth of a unit)
...
tanα=12/5
α=arctan2.4°
α≈67.38° (to nearest hundredth of a degree)
...
tan13=x/24
x=24tan13 units
x≈5.54 units (to nearest hundredth of a unit)
...
sin20=10/x
x=10/sin20 units
x≈29.24 units (to nearest hundredth of a unit)
We know that
<span>the regular hexagon can be divided into 6 equilateral triangles
</span>
area of one equilateral triangle=s²*√3/4
for s=3 in
area of one equilateral triangle=9*√3/4 in²
area of a circle=pi*r²
in this problem the radius is equal to the side of a regular hexagon
r=3 in
area of the circle=pi*3²-----> 9*pi in²
we divide that area into 6 equal parts------> 9*pi/6----> 3*pi/2 in²
the area of a segment formed by a side of the hexagon and the circle is equal to <span>1/6 of the area of the circle minus the area of 1 equilateral triangle
</span>so
[ (3/2)*pi in²-(9/4)*√3 in²]
the answer is
[ (3/2)*pi in²-(9/4)*√3 in²]
