Emergency! Exams! 50, urgent! Exams!

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The weight of each box of screws satisfies a normal distribution, the mean of which is: , the standard deviation is, and then the variance is.

You can write the distribution function of the weight of each box of screws, and according to the law of normal distribution, you can find the probability that the weight of a box of screws of this type exceeds kilograms.

Since I don't have a table at hand, I can't give you an accurate evaluation, and there are detailed examples in this book.

4000-2430-(820-620)-4000*15% advertising-(600-75)-20 hospitality-98-(360-280)+280-(160-30-50)+7-300*2%union hood wheel source +45-300*14%welfareTongli +7-300*education expenses = 4000-2430-200-600-525-20-98-80+280-80+7-6+45-42+ thousand.

15 lim(x->2)∫[2,x] f(t)dt/(1-4/x^2)

lim(x->2)f(x)/(8/x^3) =f(2)

16 lim(x->0) xsin(1/x)

1<=sin(1/x)<=1

lim(x->0)-x <= lim(x->0) xsin(1/x) <= lim(x->0)x

lim(x->0)xsin(1/x)=0

lim(x->0)(x+a)=0 a=0

17 f(a^(1/3)-1)=a-1

a^(1/3)-1)(a^(2/3)+a^(1/3)+1)

a^(1/3)-1)[(a^(1/3)-1)^2-3a^(1/3)+3-3]

a^(1/3)-1)^3-3(a^(1/3)-1)^2-3(a^(1/3)-1)

f(x)=x^3-3x^2-3x

1 s=∫[0,π/2] (2-cosx)dx= 2x+sinx|[0,π/2]=π+1

2 to a(-1,1) b(2,2).

s=∫[-1,2] (x+2-x^2) dx=(1/2)x^2+2x-(1/3)x^3|[-1,2]=4/3+3/2+1/3=5/3+3/2=19/6

3 y'=e^x y'|x=0 =1 y=x+1

y=e x and y=x+1 intersect a(0,1) x>0 e x>x+1 without solution.

4 xy=1 and x=1 intersect a(1,1).

and x=2 intersect b(2,1 2).

ds=2πydx=(2π/x)dx

v=∫[1,2] x*(2π/x)dx=3*2π=6π

1y=ln(1+x)-(x-(1/2)x^2))

y'=(1/(1+x))-1+x=x^2/(1+x) x>0,y'>0

y>=0

2 y=e^x-(1+x)

y'=e^x-1 x>0,y'>0

y>=0

3 y=e^x-ex

y'=e^x-e x>1,y'>0

y>0