We want v alone, so subtract w from both sides to move it over
This leaves 5/9 v = z - w
Next, eliminate the fraction multiplying v by multiplying both sides by its reciprocal
The reciprocal of 5/9 is 9/5
(5/9) × (9/5) cancel each other, leaving v isolated on the left
The right-hand side becomes 9/5 (z - w)
Combine to get:
v = 9/5 (z - w)
There you go!:D
3 feet make one yard.
Multiplying 3 by 10 gives 30 feet.
Dividing 30 by 7 gives approximately 4.2.
Answer: 4 ropes can be cut.
Since partial ropes cannot be counted, only 4 full ropes are possible.
Answer:
The final result is $88
Step-by-step explanation:
The Math.random () method produces a floating-point, pseudo-random number ranging from 0 to 1 (including 0 but excluding 1), which is distributed fairly evenly within that interval — allowing you to scale it to fit your preferred range.
To begin with, consider a straightforward hidden Markov model (HMM). We observe a series of outcomes from rolling a four-sided die at an "occasionally dishonest casino". At time t, the result x_t belongs to the set {1, 2, 3, 4}. The casino can either be in state z_t belonging to {1, 2}. When z_t is equal to 1, it uses a fair die, whereas when z_t is equal to 2, the die is biased towards rolling a 1. Specifically: p (x_t = 1 | z_t = 1) = p (x_t = 2 | z_t = 1) = p (x_t = 3 | z_t = 1) = p (x_t = 4 | z_t = 1) = 0.25, p (x_t = 1 | z_t = 2) = 0.7, and p (x_t = 2 | z_t = 2) = p (x_t = 3 | z_t = 2) = p (x_t = 4 | z_t = 2) = 0.1. Assume there is an equal likelihood of starting in either state at time t = 1, which leads to p (z1 = 1) = p (z1 = 2) = 0.5. The casino generally maintains the same die for several iterations, but it occasionally switches states with these probabilities: p (z_t + 1 = 1 | z_t = 1) = 0.8 and p (z_t + 1 = 2 | z_t = 1) = 0.2; likewise, p (z_t + 1 = 2 | z_t = 2) = 0.1 and p (z_t + 1 = 1 | z_t = 2) = 0.9. To find the probability p (z1 = z2 = z3) that the same die is used across the first three rolls under the HMM generative model, consider the following. If we assume the first die is state 1, the probability can be calculated as p(z1=1)=0.5, and consequently, p(z2=1|z1=1)=0.8 signifies that the same die might still be in use. Alternatively, if we start with the die in state 2, p(z1=2)=0.5 and p(z2=2|z1=2)=0.9 also provides a probability. Adjacent transition probabilities can be expressed as follows: p(z_t+1=2|z_t=1)=1-p(z_t+1=1|z_t=1)=0.2 and p(z_t+1=1|z_t=2)=1-p(z_t+1=2|z_t=2)=0.1. The equation for p(z3=1|z1=1) can thus be derived as a combination of previous probabilities: [p(z3=1|z2=2)*p(z2=2|z1=1)] + [p(z3=1|z2=1)*p(z2=1|z1=1)]=0.1*0.2+0.8*0.8=0.66. Similarly for p(z3=2|z1=2): [p(z3=2|z2=2)*p(z2=2|z1=2)]+[p(z3=2|z2=1)*p(z2=1|z1=2)]=0.9*0.9+0.2*0.1=0.83. Consequently, the overall probability for using the same die for the initial three rolls can be computed via: {p(z1=1)*p(z3=1|z1=1)}*{p(z1=2)*p(z3=2|z1=2)} = 0.5*0.66+0.5*0.83 = 0.745; thus, the probability amounts to 0.745.
The response is $12.50. This might not be correct, but it's worth testing.
