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Vilka [71]
3 years ago
6

The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds

to the indicated probability.The probability of no more than 35 defective CDs.The probability of no more than 35 defective CDs.
Mathematics
1 answer:
Lerok [7]3 years ago
4 0

Answer:

P_B(X\leq35)=P_N(X

Step-by-step explanation:

If we approximate the binomial distribution with a normal distribution, we have to apply a correction factor for the fact that we are now dealing with a continuous variable instead of a discrete one, as it was with the binomial distribution.

The probability of no more than 35 defective CDs: P(X<35)

In this case, as X=35 is not included in the interval, we start the interval from X=35-0.5=34.5.

P_B(X\leq35)=P_N(X

being Pb the probability under the binomial distribution and Pn the probability under the normal distribution.

The area for the normal distribution is the one below X=34 (or P(X<34)).

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Expand the polynomial:<br> (5a+1/5 b)^2
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Expand the following:
(5 a + b/5)^2

(5 a + b/5) (5 a + b/5) = (5 a) (5 a) + (5 a) (b/5) + (b/5) (5 a) + (b/5) (b/5):
5×5 a a + (5 a b)/5 + (5 b a)/5 + (b b)/(5×5)

(5 a b)/5 = 5/5×a b = a b:
5×5 a a + a b + (5 b a)/5 + (b b)/(5×5)

(b×5 a)/5 = 5/5×b a = b a:
5×5 a a + a b + b a + (b b)/(5×5)

Combine powers. (b b)/(5×5) = (b^(1 + 1))/(5×5):
5×5 a a + a b + b a + (b^(1 + 1))/(5×5)

1 + 1 = 2:
5×5 a a + a b + b a + (b^2/5)/5

5 a×5 a = 5×5 a^2:
5×5 a^2 + a b + b a + (b^2/5)/5

5×5 = 25:
Answer:  25 a^2 + a b + b a + (b^2/5)/5
3 0
3 years ago
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A triangle is classified as an equilateral triangle when it has
Doss [256]
C. three sides with the same length
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. Use Lagrange multipliers to find the maximum and minimum values of the function, f, subject to the given constraint, g. (Place
zzz [600]

Answer:

Minimum value of f(x, y, z) = (1/3)

Step-by-step explanation:

f(x, y, z) = x⁴ + y⁴ + z⁴

We're to maximize and minimize this function subject to the constraint that

g(x, y, z) = x² + y² + z² = 1

The constraint can be rewritten as

x² + y² + z² - 1 = 0

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x, y and z

L(x,y,z) = x⁴ + y⁴ + z⁴ - λ(x² + y² + z² - 1)

We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points, each of the partial derivatives is equal to 0.

(∂L/∂x) = 4x³ - λx = 0

λ = 4x² (eqn 1)

(∂L/∂y) = 4y³ - λy = 0

λ = 4y² (eqn 2)

(∂L/∂z) = 4z³ - λz = 0

λ = 4z² (eqn 3)

(∂L/∂λ) = x² + y² + z² - 1 = 0 (eqn 4)

We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z

4x² = 4y²

4x² - 4y² = 0

(2x - 2y)(2x + 2y) = 0

x = y or x = -y

Also,

4x² = 4z²

4x² - 4z² = 0

(2x - 2z) (2x + 2z) = 0

x = z or x = -z

when x = y, x = z

when x = -y, x = -z

Hence, at the point where the box has maximum and minimal area,

x = y = z

And

x = -y = -z

Putting these into the constraint equation or the solution of the fourth partial derivative,

x² + y² + z² = 1

x = y = z

x² + x² + x² = 1

3x² = 1

x = √(1/3)

x = y = z = √(1/3)

when x = -y = -z

x² + y² + z² = 1

x² + x² + x² = 1

3x² = 1

x = √(1/3)

y = z = -√(1/3)

Inserting these into the function f(x,y,z)

f(x, y, z) = x⁴ + y⁴ + z⁴

We know that the two types of answers for x, y and z both resulting the same quantity

√(1/3)

f(x, y, z) = x⁴ + y⁴ + z⁴

f(x, y, z) = (√(1/3)⁴ + (√(1/3)⁴ + (√(1/3)⁴

f(x, y, z) = 3 × (1/9) = (1/3).

We know this point is a minimum point because when the values of x, y and z at turning points are inserted into the second derivatives, all the answers are positive! Indicating that this points obtained are

S = (1/3)

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