Answer:
To find the average, divide 16 1/2 by 5.
Converting 16 1/2 to decimal form gives 16.5, and when divided by 5, the outcome is 3.3, which can also be represented as 3 and 3/10 in fractional form.
Thus, her average workday lasted 3.3 hours.
If logarithms confuse you, here’s a simpler way to approach this problem: What power of 2 equals 128? We know that 2^3 equals 8, thus 2^4 must equal 16. Continuing to double each time, we find that 2^5 equals 32, 2^6 equals 64, and 2^7 equals 128. Hence, the value of log₂128 is 7.
Answer: 140
Oleg, Sasha, and Dima divided 600 toys among themselves. Sasha possessed twice as many toys as Oleg, while Dima had 40 more toys than Oleg.
Let Oleg's total toys be represented by x.
Let Sasha’s toys be denoted as y.
Let Dima's toys be referred to as z.
Sasha has double the amount of toys compared to Oleg.
This translates to y = 2*x= 2x.
Dima has 40 toys greater than Oleg.
Which means z = x + 40.
The total toys shared among Oleg, Sasha, and Dima is 600.
x + y + z= 600.
Substituting y=2x and z=x+40 into the equation:
Substituting values for y and z in:
Subtracting 40 from both sides:
Dividing everything by 4:
This shows Oleg has 140 toys.
Step-by-step explanation:
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.
