Solution to problem 14.6.10 from the collection of Kepe O.E.

Problem 14.6.10 from the collection of Kepe O.?. is formulated as follows:

"Two circles are given on the plane with centers at points O1 and O2 and radii R1 and R2, respectively (R1

The solution to this problem is as follows. First, let's draw a straight line passing through the centers of these circles. Let this line intersect the outer common tangent of the circles at point T. Then the distance between the centers of the circles is equal to R2 - R1, and the distance between the point T and the center of the circle O1 (O2) is equal to R1 + r (R2 + r). Thus we get two equations:

R2 - R1 = (R1 + r) + (R2 + r) R2 - R1 = (R2 + r) - (R1 + r)

Solving these equations, we obtain the value of r:

r = (R2 - R1) / 2

Thus, we have found the radius of the circle tangent to both given circles. To construct such a circle, it is necessary to draw a circle with a center at point T and radius r.

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Problem 14.6.10 from the collection of Kepe O.?. refers to the section "Probability Theory and Mathematical Statistics" and is formulated as follows:

"Two groups of students are given. In the first group, 60% of the students passed the exam, in the second group - 75%. It is known that the first group makes up 40% of the total number of students. Find the probability that a randomly selected student passed the exam."

To solve the problem, you need to use the total probability formula and Bayes' formula. First, let's find the probability of passing the exam in the general case, using the total probability formula:

P(passed) = P(passed|1st group) * P(1st group) + P(passed|2nd group) * P(2nd group)

where P(pass|1st group) = 0.6 - probability of passing the exam in the first group, P(1st group) = 0.4 - probability of choosing a student from the first group, P(pass|2nd group) = 0.75 - probability of passing the exam in the second group, P(2nd group) = 0.6 - probability of choosing a student from the second group.

Substituting the values, we get:

P(passed) = 0.60.4 + 0.750.6 = 0.69

Now we can find the probability that a randomly selected student passed the exam while in the first group using Bayes' formula:

P(1 group|passed) = P(passed|1 group) * P(1 group) / P(passed)

Substituting the values, we get:

P(group 1|passed) = 0.6*0.4 / 0.69 ≈ 0.348

Thus, the probability that a randomly selected student from the first group passed the exam is approximately 0.348 or 34.8%.

The product in this case is the solution to problem 14.6.10 from the collection of Kepe O.?.

The problem states that the body rotates around the vertical axis Oz under the action of a pair of forces with a moment M = 16t. It is also known that at time t = 3 s the angular velocity is ? = 2 rad/s, and at t = 0 the body was at rest.

It is necessary to determine the moment of inertia of the body relative to the Oz axis.

To solve the problem, you can use the equation of dynamics of rotational motion: M = Iα, where M is the moment of force, I is the moment of inertia of the body, α is the angular acceleration.

From the conditions of the problem we know the value of the moment of force M and angular velocity ? at a certain time value t. Since at time t = 3 the body has an angular velocity? = 2 rad/s, you can find the angular acceleration α as follows: α = Δ?/Δt = (? - ?0)/(t - t0) = (2 - 0)/(3 - 0) = 2/3 rad/s².

Using the equation M = Iα and the known value of M, we can find the moment of inertia of the body: I = М/α = 16t/(2/3) = 24t.

At t = 3 we obtain the value of the moment of inertia I = 24*3 = 72.

Thus, the answer to problem 14.6.10 from the collection of Kepe O.?. is 36.

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