To solve this problem,lets say that
X = the weight of the machine components. <span>
<span>X is normally distributed with mean=8.5 and sd=0.09
We need to find x1 and x2 such that
P(X<x1)=0.03 and P(X>x2)=0.03
<span>Standardizing:
<span>P( Z< (x1 - 8.5)/0.09 ) =0.03
P(Z > (x2 - 8.5)/0.09 ) =0.03.
<span>From the Z standard table, we can see that approximately P
= 0.03 is achieved when Z equals to:</span></span></span></span></span>
<span>z = -1.88 and z= 1.88</span>
Therefore,
P(Z<-1.88)=0.03 and P(Z>1.88)=0.03 <span>
So,
(x1 - 8.5)/0.09 = -1.88 and
(x2 - 8.5)/0.09 =1.88
Solving for x1 and x2:
<span>x1=-1.88(0.09) + 8.5 and
<span>x2=1.88(0.09) + 8.5
<span>Which yields:
<span><span>x1 = 8.33 g</span>
<span>x2 = 8.67 g</span></span></span></span></span></span>
<span>Answer: The bottom 3 is separated by the weight
8.33 g and the top 3 by the weight 8.67 g.</span>
The total is 1164, I multiplied each cost by 2 and added all the results.
Answer:
10%
Step-by-step explanation:
Since the question is not complete, I will be solving for probability of 3 9s
To solve this, we would be using the principle of combination, and thus
A deck of 52 hours has 4 nines, the probability that it contains at least 3 9s is mathematically represented by the equation.
P ( x ≥ 3 ) = ( 48 C 10 · 4 C 3 ) / ( 52 C 13 ) + ( 48 C 9 · 4 C 4 ) / ( 52 C 13 )
Remember I told us were going to be using the principle of combination. Well then
nCr = n!/r!(n - r)!
Now, applying the above stated formula to the equation, we have
48C10 = 48!/10!(38!)
4C3 = 4!/3!(1!)
52C13 = 52!/13!(39!)
48C9 = 48!/9!(39!)
4C4 = 4!/4!(0!)
Solving for each independently, we find that
= 0.0412 + 0.588 = 0.1
1 is the value of a ther is always a imaginary 1 infront of a letter
