Answer:
78.5%
Step-by-step explanation:
The area of the square is equal to diameter squared
i.e 4 in × 4 in = 16 square inches
The area of the circle is π × r²
i.e π × 2² = 12.56637061
The percent of the time the point will be in the circle = (Area of the circle) ÷ (Area of the square) × 100
i.e 
× 100 = 78.53981631%
And is equal to 78.5% (rounded up to the nearest tenth of a percent)
Answer:
<em>The PIN is 1147</em>
Step-by-step explanation:
<u>Prime numbers</u>
These are integer numbers that only divide exactly by themselves and by 1. To find the first possible prime number between 1,000 and 1,500, calculate the square root of 1,000.
Thus, the first prime to consider is 31. Write the consecutive primes from 31:
31 , 37 , 41 , 43
We can see the only two consecutive primes whose product lies in the given interval are 31 and 37. Thus, the PIN is 31*37=1147
Answer:
13 grados es igual a 0.226893 radianes
Step-by-step explanation:
So first, I would assign the small van a variable for how many people it can hold, or x.
So since it makes two trips,
2x
And since the other van can hold six more people,
2x + x + 6
And 57 people are transported:
3x + 6 = 57
We can then solve.
3x = 57 - 6
3x = 51
x = 17
So since we now know a small can hold 17 people, and we know that the larger van has 6 more seats, we can just say
17 + 6 = 23
So the large van has 23 seats.
Hi there what you need is lagrange multipliers for constrained minimisation. It works like this,
V(X)=α2σ2X¯1+β2\sigma2X¯2
Now we want to minimise this subject to α+β=1 or α−β−1=0.
We proceed by writing a function of alpha and beta (the paramters you want to change to minimse the variance of X, but we also introduce another parameter that multiplies the sum to zero constraint. Thus we want to minimise
f(α,β,λ)=α2σ2X¯1+β2σ2X¯2+λ(\alpha−β−1).
We partially differentiate this function w.r.t each parameter and set each partial derivative equal to zero. This gives;
∂f∂α=2ασ2X¯1+λ=0
∂f∂β=2βσ2X¯2+λ=0
∂f∂λ=α+β−1=0
Setting the first two partial derivatives equal we get
α=βσ2X¯2σ2X¯1
Substituting 1−α into this expression for beta and re-arranging for alpha gives the result for alpha. Repeating the same steps but isolating beta gives the beta result.
Lagrange multipliers and constrained minimisation crop up often in stats problems. I hope this helps!And gosh that was a lot to type!xd
