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There are 6 cases, and 2 of the conditions are met, so the probability is 1 3
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The first course: 1) the probability that all 3 pieces are first-class:
p1=(50/80)*(49/79)*(48/78)=245/1027
2) 2 pieces are first-class products and 1 piece is second-class products
p2=(50/80)*(49/79)*(20/78)*3=1225/4108
3) The probability of one piece each of the first, second, and third class.
p3=(50/80)*(20/79)*(10/80)*6=125/6162
The second way: the basic events are: (upper, middle, lower); (upper, lower, middle); (middle, top, bottom);
Middle, bottom, top); (bottom, top, middle); (lower, middle, upper) a total of 6, meet the requirements of (upper, middle, lower) and (lower, middle, upper) a total of 2.
So the probability is p=2 6=1 3
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Set the strings c(a,b),d(c,d) m(
Then the pure bucket punch has 2x=a+c, 2y=b+d
a2+b2/2=1
c2+d2/2=1
Subtract the two formulas. 2x(a-c)=-y(b-d)
cd slope k=(a-c) pin loss(b-d)=-2x y (3)cd point(2,1) k=(y-1) do j(x-2) (4)3)=(4).
y(y-1)=-2x(x-2)
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a^2=52/5+1/b^2=1
b^2=5/3
x^2/5+y^2/(5/3)=1
Let the orange ridge rent a(x1,y1)b(x2,y2)x1-x2)(-1 2*2)+3(y1-y2)2y=0 slope circle trillion k=(y1-y2) (x1-x2)=1 (6y)=(y-0) (1 wild liter2-(-1)).
y^2=1/12
k^2=1/3
x1+x2=-1
y1+y2=-2
ab = 2 root number 3
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1. The second year of high school is the average of the three grades; The sample size is also the average of the three grades; i.e. 360 3 = 120 (person).
So the third year of high school is 360 100 120 140 (people);
2. The probability is: 1 2 * 1 2 * (1-1 2) * 3 = 3 8
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First of all, it depends on how many options each student has
Participate in a group, 4 situations; Participate in two groups, 6 situations, so there are 10 situations in total.
Then 42 students were assigned to these 10 situations, and at least 11 students were in exactly the same situation
Hope it helps the landlord
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Solution: From the problem, each person should participate in at least one group.
There are 42 students in the class, in four groups.
42 4 10 people. 2 people.
At least 10 students participate in the same extracurricular group.
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,b=-1/2
ax+by)^2+(ay-bx)^2=(a^2+b^2)(x^2+y^2)=34
a=-3,b=7,c=-2
a-b) 2+(b-c) 2+(c-a) 2=3(a 2+b 2+c 2)-(a+b+c) 2When a+b+c=0 (there are multiple cases), the maximum value is 2008
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<> the later draft, the state ode continues to be keyed, Zheng Jijin.
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Such as a smile and a rampant dust, trying to destroy the sedan car.
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<> watched Sun Xiao talk about serving the plan.
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Please take a clear picture.
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Let the common ratio be q, then there is.
a5=a3*q 2, q 2 >0, so a5 and a3 have the same name (either they are positive or negative), and the game laughs from a5+a3>0, knowing that a5 and a3 are both macro positive numbers.
a1*a5 = a3 2 = 4, so a3 = 2, from a3 + a5 = 8 a5 = 6;
Substituting the formula a5 = a3*q 2, q2 = 3
a13/a9 = q^4 = 9
Guo Dun: The average interval between 100 ships arriving at the port (hours) = 1182 100=, the number of ships arriving at the port within 1 year (ships) = 365 24, the average loading and unloading time of 100 ships (hours) = 1120 100=, the ** unloading time (hours) = 741 of ships arriving at the port within 1 year, the maximum number of ships that can be unloaded in 1 year (ships) = 741, the total detention time of ships (hours) = 741 in 1 year, The total ship detention fee payable by the port within 1 year (yuan) = 5600 100 = 560 000, and the condition given is about 600,000 yuan. >>>More
2. In order to have the smallest natural value, the number of digits must be as small as possible. >>>More
If two circles intersect at the points a(1,3) and b(m,-1), and the centers of the two circles are on the line x-y+c=0, what is the value of m+c? To write out the detailed process!! >>>More
Let the first disc have x pcs.
x+4 is the same number, a total of 4 plates, 100 sugars, so 4 (x+4) = 100 >>>More