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Shalnov [3]
3 years ago
5

The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 89 inches comma89 inches, and

a standard deviation of 12 inches.12 inches. What is the probability that the mean annual snowfall during 3636 randomly picked years will exceed 91.8 inches question mark91.8 inches? Round your answer to four decimal places.
Mathematics
1 answer:
polet [3.4K]3 years ago
4 0
sorry wish I could help but iI forgot all this:(
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Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +
Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

<u>Region inside the quarter disc:</u>

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

<u>Region at the limit of the disc:</u>

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

<u>Analyse in g2:</u>

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

<u>Analyse in g3:</u>

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

<u>Analyse in g1:</u>

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

\lambda_{1}=\sqrt{1/2}*-1 =-0.29

\lambda_{2}=-\sqrt{1/2}*-1 =-1.71

\lambda_{2} give us (x,y) values negatives, outside the region, so we do not take it in account

For \lambda_{1}: S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

<u>Points limits between g1, g2 y g3</u>

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

<u>Conclusion:</u>

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0
3 years ago
What is the volume of this aquarium 24 in 8 in 12 in
mote1985 [20]
2304 cubic inches. You multiply the length, width, and base
4 0
3 years ago
Read 2 more answers
find the distance between each pair of points. if necessary, round to the nearest tenth. a(0,3), b(0,12)
statuscvo [17]
The distance between a & b is 3
7 0
3 years ago
Please help! Correct answer only!
Tcecarenko [31]

Answer:

<em>" Expected Payoff " ⇒ $ 1.56 ; Type in 1.56</em>

Step-by-step explanation:

Consider the steps below;

Tickets That Can Be Entered - 1 Ticket,\\Total Tickets Entered - 1000 Tickets,\\\\Proportion - 1 / 1000,\\Money One ( First Ticket ) = 820 Dollars,\\Money One ( Second Ticket ) = 740 Dollars,\\\\Proportionality ( First ) - 1 / 1000 = x / 820,\\Proportionality ( Second ) - 1 / 1000 = x / 740\\\\1 / 1000 = x / 820,\\1000 * x = 820,\\x = 820 / 1000,\\x = 0.82,\\\\1 / 1000 = x / 740,\\1000 * x = 740,\\x = 740 / 1000,\\x = 0.74\\\\Conclusion ; " Expected Payoff " = 0.82 + 0.74 = 1.56

<em>Solution; " Expected Payoff " ⇒ $ 1.56</em>

4 0
3 years ago
Insert grouping symbols in 5 x 2 + 8 - 4 - 2 so that its value is 15
Arlecino [84]
5×2+8-4+1=15
cause
5×2=10
10+8=18
18-4=14
14+1=15
5 0
3 years ago
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