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.
Just substract corresponding terms
Option B
2(x-3)=7x-31
2x-6=7x-31
-7x -7x
----------------
-5x-6=-31
+6 +6
--------------
-5x=-25
Divide by -5
x=5
3x-2(x+5)=-3+18
15
3x-2(x+5)=15
3x-2x-10=15
x-10=15
+10 +10
--------------
x=25
F(x) = x/4
so
f(-8) = -8/4 = -2
answer
f(-8) = -2