Answer: Stephens dog weighs 60 pounds
Step-by-step explanation:
Answer:
Infinitely many solutions.
Step-by-step explanation:
Let's solve your system by elimination.
x−3y=9;−x+3y=−9
x−3y=9
−x+3y=−9
Add these equations to eliminate x:
0=0
<u>Answer:</u>
Infinitely many solutions.
7 2/9 < 7.228 < 29/4 < 7.60
Answer:
Transitive property of equality
Step-by-step explanation:
By the definition of transitivity, a relation R is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
If p = q, q = r then p = r.
Here, we have given that if ZXY = FDE and FDE = CAB, then, ZXY = CAB.
Therefore, it shows the transitive property of equality.
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
