Answer:
a.P<x<9.85 orx>10.15)=0.3174, Total defects=317.4
b.p=0.0026,total defects=2.6
c.Less of the items produced will be classified as defects.
Step-by-step explanation:
a.The standard score,z, is esentially x reduced by process mean then divided my process standard deviation.
Total defects=Production*Probability
=0.3174*1000
b. 
therefore:-
Defects=Probability*Production
=0.0026*1000
=2.6
c.Reducing process variation results in a significant reduction in the number of unit defects.
The answer would be 0
Reminder: when you see “-(-“ always think of plus sign “+.” (My teacher in middle school taught me that). So, -3+3 would be that the answer is 0.
Answer:
B. feet
Step-by-step explanation:
With the help of the given equation, we know that the automobile is worth $12528.15 after four years.
<h3>
What are equations?</h3>
- A mathematical equation is a formula that uses the equals sign to represent the equality of two expressions.
- a formula that expresses the connection between two expressions on each side of a sign.
- Typically, it has a single variable and an equal sign.
- Like this: 2x - 4 Equals 2.
- In the above example, the variable x exists.
So, the equation of depreciation: y = A(1 - r)∧t
The current value is y.
A is the initial cost.
r is the depreciation rate.
t is the time in years, and
In four years, we must ascertain the present value.
Now,
y = $24000(1 - 0.15)⁴
y = 24000(0.85)⁴
y = 24000 × 0.52200625
y = 12528.15
Therefore, with the help of the given equation, we know that the automobile is worth $12528.15 after four years.
Know more about equations here:
brainly.com/question/28937794
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Complete question:
The general equation for depreciation is given by y = A(1 – r)t, where y = current value, A = original cost, r = rate of depreciation, and t = time, in years. The original value of a car is $24,000. It depreciates 15% annually. What is its value in 4 years? $
Answer:
log5(125)=3 The 5 should be smaller and a little lower.
Step-by-step explanation:
