First we'll consider the floor size.
In a 10 by 12 room the longest pipe that will fit lying on the floor equals the hypotenuse of a 10 by 12 triangle:
hypotenuse² = 10² + 12²
hypotenuse² = 100 + 144
hypotenuse² = 244
hypotenuse =
<span>
<span>
<span>
15.62 feet
Now we have to consider the third dimension (height = 8 feet).
The two sides would be 15.62 and 8
</span></span></span>hypotenuse² =
15.62² + 8²
hypotenuse² =
244 + 64 = 308
hypotenuse = 17.55
So, the longest pipe that would fit in that room would be 17.55 feet long.
The standard form is
(x-xo)² + (y-yo)² = r²
Where xo,yo is the center of the circumference
and r is the radius.
We find the distance between the points to determine the <span>diameter
d=sqrt ((9-2)</span>² + (4-4)²<span>) = sqrt (49) = 7
The radius of the circumference is d/2 = 7/2 = 3.5
And we can see that those points are at the same coordinate y = 4
The center is given then by , x = (9+2)/2 = 5.5
y = (4+4)/2 = 4
The equation results
(x-5.5)</span>² + (y-4)² = (3.5)²<span>
</span>
Answer: 2
Step-by-step explanation:
A dice has 6 numbers on it, 1,2,3,4,5, and 6.
how many pairs of 3 can you make?
(1 2 3) (4 5 6)
There are 2 pairs, so your answer is 2
Answer
a) y | p(y)
25 | 0.8
100 | 0.15
300 | 0.05
E(y) = ∑ y . p(y)
E(y) = 25 × 0.8 + 100 × 0.15 + 300 × 0.05
E(y) = 50
average class size equal to E(y) = 50
b) y | p(y)
25 | 
100 | 
300 | 
E(y) = ∑ y . p(y)
E(y) = 25 × 0.4 + 100 × 0.3 + 300 × 0.3
E(y) = 130
average class size equal to E(y) = 130
c) Average Student in the class in a school = 50
Average student at the school has student = 130
