Answer: the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
Step-by-step explanation:
Given data;
lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min
hence, height = 1 / ( 52.0 - 50.0) = 1 / 2
now the probability that the class length is between 50.8 and 51 min = ?
P( 50.8 < X < 51 ) = base × height
= ( 51 - 50.8) × 1/2
= 0.2 × 0.5
= 0.1 ≈ 10%
therefore the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
Answer:
Step-by-step explanation:
Let n = the smaller of the two numbers, and since the other number is 5 more than twice the smaller number n, then ...
Let 2n + 5 = the second and larger number.
Since the sum of the two unknown numbers is 26, then we can write the following equation to be solved for n as follows:
n + (2n + 5) = 26
n + 2n + 5 = 26
Collecting like-terms on the left, we get:
3n + 5 = 26
3n + 5 - 5 = 26 - 5
3n + 0 = 21
3n = 21
(3n)/3 = 21/3
(3/3)n = 21/3
(1)n = 7
n = 7
Therefore, ...
2n + 5 = 2(7) + 5
= 14 + 5
= 19
CHECK:
n + (2n + 5) = 26
7 + (19) = 26
7 + 19 = 26
26 = 26
Therefore, the two desired numbers whose sum is 26 are indeed 7 and 19.
Y is the independent variables
k is the dependent variable
Answer:
Total surface area : 733
The shape of the base is a rectangle with sides 11 in. and 12 in.
Step-by-step explanation:
The shape of the base is a rectangle with sides 11 in. and 12 in.
The surface area is the sum of the areas of the 5 sides.
Area of the base = 11*12 = 132
Area of the two triangles = (11*16)/2 = 88
Area of the back rectangle = 192
The theorem of Pitagora to find the oblique side: square root of (11*11 + 16*16)= 19.42 in.
So the area of the oblique face: 19.42* 12 = 233 (almost :) )
So total surface area: 132 + 88*2+192+233= 733 square in
It’s 10077 because it’s hey
