so, we have two 54x18 rectangles, so their perimeter is simply all those units added together, 54+54+54+54+18+18+18+18 = 288.
we know the circle's diameter is 1.5 times the width, well, the width is 18, so the diameter of the circle must be 1.5*18 = 27.
![\bf \stackrel{\textit{circumference of a circle}}{C=d\pi }~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ d=27 \end{cases}\implies C=27\pi \implies C=\stackrel{\pi =3.14}{84.78} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perimeter of the rectangles}}{288}~~~~+~~~~\stackrel{\textit{perimeter of the circle}}{84.78}~~~~=~~~~372.78](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bcircumference%20of%20a%20circle%7D%7D%7BC%3Dd%5Cpi%20%7D~~%20%5Cbegin%7Bcases%7D%20d%3Ddiameter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20d%3D27%20%5Cend%7Bcases%7D%5Cimplies%20C%3D27%5Cpi%20%5Cimplies%20C%3D%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B84.78%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20rectangles%7D%7D%7B288%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20circle%7D%7D%7B84.78%7D~~~~%3D~~~~372.78)
<span>1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?
25% chance
</span><span>2. Describe how to use two coins to simulate which two pets the family has.
</span>
You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).
<span>3. Flip both coins 50 times and record your data in a table like the one below.
</span><span>Based on your data, what is the experimental probability that the family has two dogs or two cats?
</span>
Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance
<span>4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
1/8 chance (accidentally messed up there) or 12.5%
</span><span>5. How could you change the simulation to generate data for three pets?
</span><span>
To flip 3 coins and add more spots on the chart.
I hope that this helps because it took a while to write out. If it does, please rate as Brainliest
</span>
Sec is 1/cos so sec -120 is 1/cos -120 which is -2
Answer:
15.59
Step-by-step explanation:
The answer is 15.59 because to find the area of an equilateral triangle, you need to multiply the height by the length. In this scenario, the length is 6, and we need to find the height. Since we know that all of the sides of the equilateral triangle are 6, we can put a line dividing the equilateral triangle exactly in half as our height. This will form 2 triangles, and each of them will have a length of 3, and a slant height of 6. They would also be right triangles. Now, with this information you can do pythageroum thereom to find the height:: 3^2 + h^2 = 6^2. So, h = √27. Now, to find the area of a triangle, you must multiply the length by the height and then divide by 2. So (6 • √27)/2 is equal to 15.5884572681, which rounded is equivalent to 15.59. So the answer is 15.59.
