5 1/8 8 goes into 41 5 times with 1 left over
Answer:
<em>Their masses are 100 kg and 90 kg</em>
Step-by-step explanation:
<u>Equations</u>
We'll solve the problem by setting only one variable. Let's call:
x = weight of the heavier boxer
Since the sum of the masses of both boxers is 190 Kg:
190 - x = mass of the lighter boxer.
It's given that
of the mass of the heavier boxer is 10 kg less than the mass of the other, thus:

Operating:

Multiplying by 5:


Simplifying:


The heavier boxer's mass is 100 kg. The lighter boxer has a mass of 190-100 = 90 kg.
Their masses are 100 kg and 90 kg
Answer:
The answer is below
Step-by-step explanation:
Let us assume that the length of the square be x. Therefore the area of the square is given as:
Area of square = length × length = x × x = x²
Given that the length of one side of the rectangle is twice the length of the other side. Let the length of the small side be a therefore the length of the long side = 2a
The area of the rectangle = length of small side × length of long side = a × 2a = 2a²
Since both the rectangle and square are to occupy the same space, hence:

Length of small side = a =
, length of long side = 2a = 
The length of the sides are length of the square side divide by √2 (
) and twice the length of the square side divide by √2 (
)
The answer is <span>mean = 13,027; median = 12,200; no mode
</span>
Let's rearrange values from the lowest to the highest:
11350, 12050, 12200, 13325, 16211
<span>The mean is the sum of all values divided by the number of values:
</span>(11350 + 12050 + 12200 + 13325 + 16211)/5 ≈ 13027
The median is the middle value. If there is an odd number of data, then the median is the value in the middle. In the data set 11350, 12050, 12200, 13325, 16211, the median (the middle value) is 12200
<span>The mode is the value that occurs most frequently. Since none of the number does not occur most frequently, there is no mode.
</span>
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