The class width for this Frequency Distribution Table is 5.
<h3>What is the class width?</h3>
The class width is the difference between the upper boundary and the lower boundary of the class.
The class width = upper boundary - lower boundary
5 - 0 = 5
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Answer:
5/9
Step-by-step explanation:
Start be letting x = 0.555...
Our original equation is:
x = 0.555...
There is only one digit repeating, the 5, so we multiply both sides of that equation by 10 and write it above the original equation.
10x = 5.555...
x = 0.555...
Now we subtract the second equation from the first equation as written above.
9x = 5
Divide both sides by 9.
x = 5/9
0.555... = 5/9
Answer:
<em>AB = 7.35 cm</em>
Step-by-step explanation:
From the attachment,
In ΔDEF,
DF = GH-(GD+FH) = 6 - (2+3) = 1 cm
DE = 2+3 = 5 cm (sum of two radius)
Applying Pythagoras theorem,
In ΔCDI,
DI = GH-(GD+IH) = 6 - (2+1.5) = 2.5 cm
CD = 2+1.5 = 3.5 cm (sum of two radius)
Applying Pythagoras theorem,
AB = EF + CI = 
Answer:
1
Step-by-step explanation:
using PEDMAS(parentheses, exponents, division, multiply, addition, subtraction) to solve the problem
First we multiply (4)(-3)
(4)(−3)−5−3(−6)
Then multiply -3(-6)
(-12)-5-3(-6)
=-12-5+18
subtract 12-5
-12-5+18
=-17+18
Adding -17+18 gives us 1.
Therefore, (4)(−3)−5−3(−6) is 1.
