In RT △ ABC, the angle c = 90 ° and if Tana = 3, then the value of SINB is

Tana = A / b = 3 / 1 and C = 1 ^ 2 + 3 ^ 2 = √ 10a, so SINB = √ 10 / 10

B and C in the quadratic equation of one variable x2 + BX + C = 0 are the number of points that appear twice in a dice roll. Find the probability that the equation has real roots and the probability that it has multiple roots

When a dice is rolled twice, the total number of basic events is 36
The number of basic events with b square > = 4C or C = 4C is 0 1 2 4 6 6
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Meet C

Let the quadratic equation of one variable x2 + BX + C = 0, if B and C are the number of points that appear twice in a dice roll, the probability that the equation has real roots can be obtained

When B = 2, C = 1; & nbsp; there is one case. When B = 3, C = 1,2; & nbsp; there are two cases. When B = 4, C = 1,2,3,4; & nbsp; there are two cases

In the quadratic equation of one variable x2-2 (A-2) x-b2 + 16 = 0, if a and B are the points obtained by randomly rolling dice, then the probability that the equation has two positive roots is zero______ ．

Because a and B are random dice points, there are 36 results. And because the quadratic equation x2-2 (A-2) x-b2 + 16 = 0 has two positive roots, so △ = 4 (A-2) 2-4 (16-b2) > 0 and X1 + x2 = 2 (A-2) > 0 and x1x2 = 16-b2 > 0, the solution is: a = 5, B = 3; a = 6, B = 1; a = 6, B = 2

Consider the univariate quadratic equation x2 + BX + C = 0, where B and C are the number of points that appear successively in two successive rolls of a dice, and find the probability p with real roots and the probability q with multiple roots of the equation

If a chromon is thrown twice, the total number of basic events is 36, and the equation has real roots if B2 ≥ 4C, it is easy to know that: B1 & nbsp; 2 & nbsp; 3 & nbsp; 4 & nbsp; 56 & nbsp; the number of basic events making C ≤ B24 & nbsp; 0 1 & nbsp; & nbsp; 2 4 & nbsp; 6 & nbsp; the number of basic events making C = B24 & nbsp; 0 1 & nbsp; 0 & nbsp; 1 & nbsp; &It can be seen from the above table that: if the equation x2 + BX + C = 0 has real roots, the total number of basic events is 1 + 2 + 4 + 6 + 6 = 19, so the probability of equation having real roots is p = 1936, and the sufficient and necessary condition of equation having multiple roots is B2 = 4C. There are two basic events satisfying this condition, so the probability of equation having multiple roots is q = 236 = 118

What is the probability that the square of the equation x + BX + C = 0 has two real roots if the number of the two chromatics is B and C respectively

The equation x & sup2; + BX + C = 0 has two real roots
∴Δ=b²-4c＞0
And B, C are the number of dice, B, C are the number of 1,2,3,4,5,6
The value of C is discussed
There are 17 kinds of (3,1), (4,1) (5,1), (6,1), (3,2), (4,2), (5,2), (6,2), (4,3), (5,3), (6,3), (5,4), (6,4), (5,5), (6,5), (5,6), (6,6), (6,6) in total
There are 36 species of B and C
The probability is 17 / 36

If a dice is tossed randomly twice, and the number of upward points is a and B respectively, then the equation AX + B = 0 about X has an integer solution, and the probability is 0

A dice is tossed randomly twice, and the number of up points is a and B respectively. There are 6 * 6 = 36 combinations
The solution of AX + B = 0 is: x = - B △ a
The number of cases with integer solutions is as follows:
When a = 1, B = 1,2,3,4,5,6 can be used
When a = 2, B = 2,4,6 three cases meet the requirements
When a = 3, B = 3, 6 two cases meet the requirements
When a is equal to 4,5,6 respectively, B is equal to 4,5,6 respectively,
There are 14 cases of 6 + 3 + 2 + 1 + 1 + 1
So that the equation AX + B = 0 about X has integer solution, the probability is: 14 / 36 ≈ 38.9%
I hope I can help you

If you take any two real numbers in the interval (0,1), the probability that the sum of the two real numbers is less than 0.8 is zero___ ．

Let two numbers be taken as X, y; then 0 & lt; X & lt; 10 & lt; Y & lt; 1, if the sum of the two numbers is less than 0.8, then 0 & lt; X & lt; 10 & lt; Y & lt; 1x + Y & lt; 0.8. According to the geometric probability, the original problem can be transformed into finding the regions 0 & lt; X & lt; 10 & lt; Y & lt; 1x + Y & lt; 0.8 and 0 & lt; X & lt; 10 & lt; Y & lt; 0; 1 represents the area ratio of the region, as shown in the figure; when the sum of the two numbers is less than 0.8, the corresponding point falls on the shadow, ∵ s shadow = 12 × 0.8 × 0.8 = 825, so if two numbers are randomly taken out in the interval (0, 1), the probability that the sum of the two numbers is less than 0.8 is p = 8251 × 1 = 0.32

If we take any two real numbers in the (0,1) interval, the probability that their sum is greater than 12 and less than 54 is zero______ ．

Let two numbers be x and Y respectively, then 0 ＜ x ＜ 10 ＜ y ＜ 1, and the area is a square with side length of 1. If the sum of them is greater than 12 and less than 54, then a: 0 ＜ x ＜ 10 ＜ y ＜ 112 ＜ x + y ＜ 54 corresponds to the shaded area shown in the figure, and its area is s = 1-s △ ebf-somn = 1-12 × 34 × 34-12 × 12 = 1932  P (a) = 1932, so the answer is: 1 nine hundred and thirty-two

If we take any real number in the interval [0,10], then the probability that the sum of it and 6 is greater than 12 is zero

X ∈ [0,10], and X + 6 > 12, that is: x > 6, so x ∈ (6,10]
Then: P = (4) / (10) = 2 / 5

In RT △ ABC, the angle c = 90 ° and if Tana = 3, then the value of SINB is