in the equation as we can say x= 2y-10.
so, substitute x=2y-10 everywhere then the x in 1240 gets cancelled with 2y-10 then solve the remaining you will get 4y-20+3y=1240
7y=1260
y=1260/7
y=180
then substitute the value of y everywhere
2x+3y=1240
2x+3(180)=1240
2x=1240-540
2x=700
x=350
DONE.
Answer:
1) distributive property.
Step-by-step explanation:
Becca used the distributive property, in that she distributed (multiplied) 4 to all terms within the parenthesis:
4(x + 2) = 4x + 8
~
Answer: The correct graph is the bottom left graph.
Step-by-step explanation:
Given function is f(x)=ceil(x+1)
To plot graph of f(x) in interval of(-3,3) :
ceil(x+1) is ceiling function
The output of ceil(x) is least integer greater than x
for example ceil(5.5)=6
For an interval of (-3,-2):
Take x=(-2.4)
x+1=(-1.4)
y=f(x)=ceil(x+1)=(-1)
Similarly,
For an interval of (-2,-1):
Take x=(-1.4)
x+1=(-0.4)
y=f(x)=ceil(x+1)=(0)
For an interval of (-1,0)
y=f(x)=1
For an interval of (0,1)
y=f(x)=2
For an interval of (1,2)
y=f(x)=3
For an interval of (2,3)
y=f(x)=4
Thus, The correct graph is the bottom left graph.
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 .
