Answer:
Step-by-step explanation: 54in
In a large population, 61% of the people are vaccinated, meaning there are 39% who are not. The problem asks for the probability that out of the 4 randomly selected people, at least one of them has been vaccinated. Therefore, we need to add all the possibilities that there could be one, two, three or four randomly selected persons who were vaccinated.
For only one person, we use P(1), same reasoning should hold for other subscripts.
P(1) = (61/100)(39/100)(39/100)(39/100) = 0.03618459
P(2) = (61/100)(61/100)(39/100)(39/100) = 0.05659641
P(3) = (61/100)(61/100)(61/100)(39/100) = 0.08852259
P(4) = (61/100)(61/100)(61/100)(61/100) = 0.13845841
Adding these probabilities, we have 0.319761. Therefore the probability of at least one person has been vaccinated out of 4 persons randomly selected is 0.32 or 32%, rounded off to the nearest hundredths.
<span>c.<span>Loan I's monthly payment will be $11.88 smaller than Loan H's.</span></span>
Answer:
- hexahedron: triangle or quadrilateral or pentagon
- icosahedron: quadrilateral or pentagon
Step-by-step explanation:
<u>Hexahedron</u>
A hexahedron has 6 faces. A <em>regular</em> hexahedron is a cube. 3 square faces meet at each vertex.
If the hexahedron is not regular, depending on how those faces are arranged, a slice near a vertex may intersect 3, 4, or 5 faces. The first attachment shows 3- and 4-edges meeting at a vertex. If those two vertices were merged, then there would be 5 edges meeting at the vertex of the resulting pentagonal pyramid.
A slice near a vertex may create a triangle, quadrilateral, or pentagon.
<u>Icosahedron</u>
An icosahedron has 20 faces. The faces of a <em>regular</em> icosahedron are all equilateral triangles. 5 triangles meet at each vertex.
If the icosahedron is not regular, depending on how the faces are arranged, a slice near the vertex may intersect from 3 to 19 faces.
A slice near a vertex may create a polygon of 3 to 19 sides..
Caleb draws 14 dogs on each of 4 posters.
Then,
Number of dogs drawn by caleb in the first instance = (14 * 4) dogs
= 56 dogs
Celeb drew 18 cats on each of 6 other posters.
Then
Number of cats drawn by Caleb in the first instance = (18 * 6) cats
= 108 cats
Again Caleb draws 5 more dogs on each poster with dogs.
Then,
The number of Dogs drwn in the second instance = (5 * 4) dogs
= 20 dogs
So, the total number of dogs drawn by Caleb = ( 56 + 20) dogs
= 76 dogs
Then the total number of cats and dogs drawn by Caleb = 108 + 76
= 184
