I am assuming you want to solve for m in each case
8n = -3m + 1
8(-2) = -3m + 1
-16 = -3m + 1
-3m = -17
m = 
8(2) = -3m + 1
16 = -3m + 1
-3m = 15
m = -5
8(4) = -3m + 1
32 = -3m + 1
-3m = 31
m = 
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.
Answer:
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Step-by-step explanation:
P(B) = 0.75.
For independent events, P(A and B) = P(A)*P(B). This gives us
1/8 = 1/6(x)
Divide both sides by 1/6:
1/8 ÷ 1/6 = x
1/8 × 6/1 = x
6/8 = x
3/4 = x
0.75 = x