Answer:
15 seats
Step-by-step explanation:
It is given that section A of a theater contains 12 rows of 15 seats each.
Now, it is also given that section B contains only 10 rows it has an equal number of seats the same as section A.
As the number of seats in each row of section A is 15, so, the number of seats in each row of section B will be 15 and the same as section A. (Answer)
Answer:
∠RST = 40°
Explanation:
We are given that:
SQ bisects angle RST.
This means that:
∠QST = ∠QSR
2x = 3x - 10
10 = 3x - 2x
x = 10
Therefore:
∠QST = 2x = 2(10) = 20°
∠QSR = 3x - 10 = 3(10) - 10 = 30 - 10 = 20°
Note that both angles are equal.
Now, we can get ∠RST as follows:
∠RST = ∠QST + ∠QSR
∠RST = 20 + 20 = 40°
Hope this helps :)
9.4 Ft for the first one
7.5 Ft for the second piece
12 Ft for the Third
he uses 5.2 Ft from the first piece of wood so 9.4 - 5.2= 4.6ft
he uses 6.1 ft from the second piece of wood , 7.5 - 6.1 = 1.4 ft
Nathan didnt use any of the 3rd piece of wood , so he still have 12 Feet of wood remaining so, 12+4.6+1.4 = 18 feet of wood left that he didnt use.
Answer:
The function (gof)(x) is;
Explanation:
Given the functions;
Solving for the function;
so, we have;
Therefore, the function (gof)(x) is;
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 .
