Answer:
There are no solutions to this question.
Step-by-step explanation:
Answer:
Bottom Right Corner
Step-by-step explanation:
The top left corner represents the Commutative Property of Addition, which says that order doesn't matter when you're adding two things together. In this case, the two items are Pretzels (p) and Mustard (m): p + m = m + p.
The bottom left corner represents the Distributive Property, which says that if you are expanding out a parentheses, you must add the products of the outside number with each of the inside numbers. In this case, the outside number if Pretzel (p) and the inside numbers are Mustard (m) and -1:
p(m - 1) = pm - p
The top right corner represents the Zero Identity Rule, which states that anything multiplied by 0 equals 0. Here the "anything" is just Pretzels (p), Salt (s), and Mustard (m): (p + s + m) * 0 = 0.
The bottom right corner simply states that you can break up a fraction into the product of the numerator and 1 over the denominator. In this case, the numerator is Pretzels (p) and the denominator is Salt (s):
. However, this is not a specific "rule" or "property". It is simply an equivalence. That's why it's the odd one out.
Thus, the bottom right corner is the one that doesn't belong.
Hope this helps!
Sometimes, for it depends on the triangle. Take a right triangle, for instance. In a right triangle, the hypotenuse is always greater than the two legs. Therefore, you cannot label any side as the base, since the sides are different. In an equilateral triangle, on the other hand, all of the sides and angles are the same. Therefore, no matter which way you turn the triangle, every side is equal. In this sense it doesn't matter which side is labeled as the base. So, it really depends on the kind of triangle you're referring to.
Statement Problem: P, Q and R are three buildings. A car began its journey at P, drove to Q, then to R and returned to P. The bearing of Q from P is 058° and R is due east of Q. PQ = 114 km and QR = 70 km. ( Draw a clearly labelled diagram to represent the above information.
Solution:
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