Answer:
An adult Welsh Corgi is always going to be shorter than an adult Basset Hound.
Step-by-step explanation:
That's what I think because the Basset Hound is 13.8 in. in height right? So he's two inches taller if we round up both numbers. Basset hound would be 14 and welsh corgi would be 11....................okay im not sure i give up but it's either
"An adult Welsh Corgi is always going to be shorter than an adult Basset Hound."
"Basset hounds tend to be a little over two inches taller than Welsh Corgis, but have a higher variability in height.
whichever just not the other ones good luck!
Answer:
The radius of the circle is 5 units
Step-by-step explanation:
Given in the question,
centered point = (-3 , 2)
a point in the circle = (1,5)
To find the radius of circle we will use the circle equation
(x – h)² + (y – k)² = r²,
with the centre being at the point (h, k) and the radius being r.
Plug values in the equation
(1 – (-3))² + (5 – 2)² = r²
(1+3)² + (3)² = r²
4² + 3² = r²
25 = r²
r = √25
r = 5
The smallest number of tiles Quintin will need in order to tile his floor is 20
The given parameters;
- number of different shapes of tiles available = 3
- area of each square shape tiles, A = 2000 cm²
- length of the floor, L = 10 m = 1000 cm
- width of the floor, W = 6 m = 600 cm
To find:
- the smallest number of tiles Quintin will need in order to tile his floor
Among the three different shapes available, total area of one is calculated as;
Area of the floor is calculated as;
The maximum number tiles needed (this will be possible if only one shape type is used)
When all the three different shape types are used we can get the smallest number of tiles needed.
The minimum or smallest number of tiles needed (this will be possible if all the 3 different shapes are used)
Thus, the smallest number of tiles Quintin will need in order to tile his floor is 20
Learn more here: brainly.com/question/13877427
The giant jar of jelly beans is 5 times the size of the smaller jar. If there are 83 jelly beans in the smaller jar, the giant jar can be represented as 5(83).
5(83) = 415 jelly beans
Answer:
81%
Step-by-step explanation:
Let 'L' be the dominant and 'l' e the recessive allele for ‘lazybuttness’.
Since ‘lazybuttness’ is an autosomal dominant condition, the 19% of students affected by the condition correspond to the homozygous dominant (LL) and heterozygous (Ll) genotypes. Therefore, the rest of the population has the homozygous recessive genotype (ll) and is not affected. The frequency of students not affected is:
F = 100% - 19% = 81%
