Answer:
Length of the arc = 5.25017 feet
Step-by-step explanation:
The line from the middle of the tree to the tip of the tangent lines is 33 ft and from the circumference of the tree its 27ft. One of the right angle triangles is picked and the angle is determined using SOHCAHTOA
Hence Tan X = 6/27
X = Tan-1 (6/27) = 12.52880770915o
Hence both sum of angles where tangent lines meet = 2 x 12.52880770915o = 25.05761541830o
Angle of the arc inside the circle = 50.115o
Length of the arc = 50.115/360 x (2x22/7 x 6) = 5.25017040025ft
1
2
3
4
5
6
Frequency
32
36
44
20
30
38
What is the experimental probability of rolling the given result?
a.
0.88
c.
0.56
b.
0.34
d.
0.44
The experimental probability of rolling the given results are: a. 0.16, b. 0.18, c. 0.22, d. 0.1. The correct answer is b. 0.18.
To find the experimental probability of rolling a given result, divide the frequency of that result by the total number of rolls. In this case, the frequency of rolling a 1 is 32, and the total number of rolls is 200. So the experimental probability of rolling a 1 is 32/200 = 0.16. Repeat this process for the other results:
Based on these calculations, the experimental probability of rolling the given results are:
Therefore, the correct answer is b. 0.18.
#SPJ2
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Explanation:
Add up all the frequencies that correspond to outcomes that are more than 3
So we'll add up the last three frequencies (since they correspond to outcomes of 4, 5 and 6 all of which are greater than 3) to get 20+30+38 = 88
We have 88 occurrences of rolling either a 4, a 5 or a 6. This is out of 200 rolls total.
The empirical (or experimental) probability of getting more than 3 on the die is 88/200 = 0.44
Exactly one tail
At least one tail
All tails
Three heads
By using sample space we can say that complement of 4 heads in the toss of 4 coins is at least one tail .
set of all possible outcomes experiment is called sample space .
To find compliment of 4 heads we will first find sample space Sample space= S
= { HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT TTHH, TTHT, TTTH, TTTT }
Given case is 4 heads Say A= { HHHH}
Now we can calculate A complement as
A' = S-A
= { HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT TTHH, TTHT, TTTH,TTTT }- { HHHH}A'={ HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT TTHH, TTHT, TTTH,TTTT }
By using sample space we can say that complement of 4 heads in the toss of 4 coins is at least one tail .
To know more about sample space visit : brainly.com/question/5750471
Answer:
b i believe the answer is
How old is Ben?
✪According to question:-
Ben is 6 years older than Ishaan
So :-
put supposed values
Original Ages :-
✪Verification :
1st condition:-
put original ages
LHS=RHS
Hence verified !
Ben's age =8years
Answer:
No, a square is NOT the cross section of a rectangular and triangular prism.
Step-by-step explanation:
Prisms have a uniform cross-section and are named after their cross-section. Hence, the cross-section of a rectangular prism is a rectangle and the cross-section of a triangular prism is a triangle. The only prism with a square cross-section is a cube.