He was wrong, you cannot make 10 coins by adding 32 coins to 28 coins. If you want to obtain 10 coins from 32 coins and 28 coins, you can subtract 28 coins from 32 coins.
This can be illustrated as follows:
32-28=10
Answer:
x = 4
Step-by-step explanation:
Simplifying
3x + 15 = 6x + 3
Reorder the terms:
15 + 3x = 6x + 3
Reorder the terms:
15 + 3x = 3 + 6x
Solving
15 + 3x = 3 + 6x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-6x' to each side of the equation.
15 + 3x + -6x = 3 + 6x + -6x
Combine like terms: 3x + -6x = -3x
15 + -3x = 3 + 6x + -6x
Combine like terms: 6x + -6x = 0
15 + -3x = 3 + 0
15 + -3x = 3
Add '-15' to each side of the equation.
15 + -15 + -3x = 3 + -15
Combine like terms: 15 + -15 = 0
0 + -3x = 3 + -15
-3x = 3 + -15
Combine like terms: 3 + -15 = -12
-3x = -12
Divide each side by '-3'.
x = 4
Simplifying
x = 4
Answer:
Step-by-step explanation:
Let
s -----> the number of stamps
w ----> the number of weeks
we know that
The linear equation that represent this situation is
----> equation of the line into slope intercept form
where
the slope m is equal to 
the y-intercept b is equal to 
---> (the initial value)
Answer and explanation:
1. It is commonly referred to as the arithmetic average: the mean measure of central tendency is also referred to as the arithmetic mean or arithmetic average or just average.
It is algebraically defined (that is, there is an equation you can use to calculate its value): the mean can be represented by the algebraic equation
a1+a2+a3...ai/n
2. There can be more than one median where there are an even number of data points and not odd number, in which the two middle number are divided by 2
It can be found when there are undetermined scores: median can be found with undetermined scores
3. It corresponds to an actual score in the data: the mode is an actual value in the data that appears more frequently than other values
There can be more than one: there can be more than one mode(bimodal,trimodal,multimodal) where 2 or more values appear the same number of times
