Answer:
2 minutes
Step-by-step explanation:
The front of the train enters the tunnel and travels 1 km before the rear of the train enters the tunnel. Thus, the total distance traveled by the front of the train is 2 km, and this is covered in a certain number of hours:
Since distance = rate times time, time = distance / rate. In this particular case we have:
2 km
time = distance / rate = ------------------- = (1/30) hr
60 km/hr
(1/30) hr converts to 2 minutes, since 1 hr = 60 minutes
Answer:
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Step-by-step explanation:
denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier
then
P(A)= ∑ P(Bi)*P(C) with i from 1 to 3
P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125
from the theorem of Bayes
P(Cz/A)= P(Cz∩A)/P(A)
where
P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained
P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100
therefore
P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Answer:
<em>The equation will be:
</em>
Step-by-step explanation:
The number of student tickets that were sold
and the number of other tickets that were sold 
Student tickets cost $3 a piece and tickets for everyone else cost $5 each.
So, <u>the total cost of</u>
<u>student tickets</u>
and <u>the total cost of</u>
<u>other tickets</u> 
Given that, the drama club sold <u>total $779 worth of tickets</u>.
So, the equation will be: 
Answer:
Yes
Step-by-step explanation:
All rhombuses are parallelograms, but not all parallelograms are rhombuses. All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent. Diagonals of a rhombus always bisect each other at right angles.
Answer:
We say that f(x) has an absolute (or global) minimum at x=c if f(x)≥f(c) f ( x ) ≥ f ( c ) for every x in the domain we are working on. We say that f(x) has a relative (or local) minimum at x=c iff(x)≥f(c) f ( x ) ≥ f ( c ) for every x in some open interval around x=c .