Answer:
- price = 380
- dollar = (price // 100) + 1
- print(dollar)
Explanation:
Presume the an item cost 380 cents and we set 380 to the variable price (Line 1). To estimate the number of single dollars to be paid, we can use // operator to divide price by 100 and we will get 3 (the remaining decimal point will be discarded). Then we add 3 by one. So the expression is
(price//100) + 1
*The reason we divide price by 100 is because 1 dollar = 100 cents
Answer:
Resonance is when the frequency of the vibrating quartz rod is equal to the frequency of the LC tank circuit.
The free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.
The first image shows the diagram of the beam and the second image shows the free-body diagram of the beam.
The resolution of forces in the system is well understood by the principle of equilibrium where a stationary body will remain balanced when subject to parallel forces provided that the total sum of the overall external forces is zero.
The free-body diagram is a graphical representation used to visualize the forces applied to an object.
The equilibrium of forces on the x-axis is:

The equilibrium of forces on the y-axis is:

The equilibrium condition at any point is:

From the free body diagram attached in the second image below,
- the horizontal reaction is located at point A as
- the vertical reaction is located at point A as
- the tension = T
- the weight = W
Therefore, we can conclude that the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.
Learn more about the free-body diagram here:
brainly.com/question/19345060?referrer=searchResults
Answer:
$$\begin{align*}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Explanation:
\eqalign{
P(Y-X=m\mid Y\gt X)
&=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr
&=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr
&=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr
}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$