Answer:
P = 0,0012 or P = 0.12 %
Step-by-step explanation:
We know for normal distribution that:
μ ± σ in that range we find 68.3 % of all values
μ ± 2σ ⇒ 95.5 % and
μ ± 3σ ⇒ 99.7 %
Fom problem statement
We have to find (approximately) % of cars that reamain in service between 71 and 83 months
65 + 6 = 71 ( μ + σ ) therefore 95.5 % of values are from 59 and up to 71 then by symmetry 95.5/2 = 49.75 of values will be above mean
Probability between 65 and 71 is 49.75 %
On the other hand 74 is a value for mean plus 1, 5 σ and
74 is the value limit for mean plus 1,5 σ and correspond to 49,85 (from z=0 or mean 65).
Then the pobabilty for 83 have to be bigger than 49.85 and smaller than 0,5 assume is 49.87
Finally the probability approximately for cars that remain in service between 71 and 83 months is : 0,4987 - 0.49.75
P = 0,0012 or P = 0.12 %
Answer:
18.0625 ft²
Step-by-step explanation:
4.25 × 4.25
18.0625 ft²
Answer:
$3.54 per pound
Step-by-step explanation:
<u>Step 1: Find the unit cost per ton</u>
28,320 / 4
<em>$7,080 per ton</em>
<u>Step 2: Find the unit cost per pound</u>
1 ton = 2000 pounds
7080 / 2000
<em>$3.54 per pound</em>
Answer: $3.54 per pound
Answer:
The answers can be seen in the photo
Answer:
a
e
f
Step-by-step explanation:
Because each time yoi divide them by two it should always become an even number.
