Answer: Well if you were going to think of a circle you can divide it into 4 equal sizes in one of the is equal to two pentagons and a half. If you add all is equal to a loop full of pentagons [which makes 10 pentagons]. Because the ring is made out of regular pentagons, we can work out that each of the interior angles of each pentagon is 108* degrees. By extending the lines that two pentagons share, assuming they will all meet in the middle, it will create a triangle. As we know each angle of a [regular] pentagon is 108* degrees, we know the two base angles of the triangle would equal to 72** which leaves the top angle to be 36 degrees. As [the sum of] angles at a point is 360 degrees, and 36 is divisible by 360, it will make a complete ring. Also, as 360 ÷ 36 = 10, we know that the ring will be made out of 10 pentagons. as formula to calculate the the size of a interior angle of a polygon is (n×180−360)÷n (for n being the number of sides that the polygon has). because the triangle is made by extending the lines, and angles on a line is 180 degrees, 180−108 (an interior angle of a pentagon) =72.
Answer:
The answer is A. (or the first one)
Step-by-step explanation:
It's not D because the median is 46. It is not B or C either because the IQR is 35 and the OQR is 53.5 and A is the only one that has that. Hope it helped :)
In this case, you aren't going to get an integer for the value of a, but you can rearrange the equation to equal a.
4a-3=D add 3 to both sides
4a=D+3 divide both sides by 4
a=(D+3)/4 here's your answer
9 yards I’m sorry if it’s wrong
Answer:
we reject H₀
Step-by-step explanation: Se annex
The test is one tail-test (greater than)
Using α = 0,05 (critical value ) from z- table we get
z(c) = 1,64
And Test hypothesis is:
H₀ Null hypothesis μ = μ₀
Hₐ Alternate hypothesis μ > μ₀
Which we need to compare with z(s) = 2,19 (from problem statement)
The annex shows z(c), z(s), rejection and acceptance regions, and as we can see z(s) > z(c) and it is in the rejection region
So base on our drawing we will reject H₀
