someone please help her or him please
Answer:
23 chalkboards
Step-by-step explanation:
Given:
Mean length = 5 m
Standard deviation = 0.01
Number of units ordered = 1000
Now,
The z factor = 
or
The z factor = 
or
Z = - 2
Now, the Probability P( length < 4.98 )
Also, From z table the p-value = 0.0228
therefore,
P( length < 4.98 ) = 0.0228
Hence, out of 1000 chalkboard ordered (0.0228 × 1000) = 23 chalkboard are likely to have lengths of under 4.98 m.
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The required range of the data is 24mm
The range of data is the difference between the largest value and the least value in a set of data.
Since we are not given the range of data. Let's assume we gave the following data:
34mm, 30mm, 20mm, 44mm, 31mm
From the data
Highest measurement = 44mm
Least measurement = 20mm
Find the range
Range = Highest measurement - Least measurement
Range = 44mm - 20mm
Range = 24mm
Hence the required range of data is 24mm
<em>NB: The concept can be applied to any set of data given for other measurements in (b) and (c)</em>
<em>Learn more here: brainly.com/question/1786006</em>
I think the answer would be B :)
