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exis [7]
3 years ago
9

1) The probability that an individual is left-handed is 12%. In a randomly selected class of 30 students, what is the probabilit

y of finding exactly 4 left-handed students?
2) Sixty-five percent of men consider themselves knowledgeable football fans. If 10 men are randomly selected, find the probability that exactly six of them will consider themselves knowledgeable fans.
Mathematics
1 answer:
svlad2 [7]3 years ago
3 0

Answer:

0.2047

0.2377

Step-by-step explanation:

1.

What we have here is a binomial distribution

Probability p = 0.12

N = number of students in the class = 30

Probability of finding 4 left handers

P(x = 4) = 30C4 x (0.12)⁴ x (0.88)²⁶

= 27405 x 0.00020736 x 0.03602047835

= 0.2047

2.

This is also a binomial distribution

N = 10 p = 0.65

P(of 6 knowledgeable fans)

= 10C6 x (0.65)⁶ x (0.35)⁴

= 210 x 0.0754189 x 0.01500625

= 0.2377

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Answer:

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A house has increased in value by 27% since it was purchased. If the current value is $254,000 , what was the value when it was
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4 years ago
3(4a+7)=5(2a+4)+1 What is it?
hichkok12 [17]

Answer:

a = 0

Step-by-step explanation:

3(4a+7)=5(2a+4)+1     Distribute 3 through the parentheses

                 - >                         12a + 21 = 5(2a + 4) +1

5(2a + 4) +1   -->  10a + 20 + 1

12a + 21 = 10a + 20 + 1  -->    12a + 21 = 10a + 21   -->      12a = 10a

                                                     12a - 10a = 0   -->    2a = 0

Divide both sides of the equation by 2      -->     a = 0

7 0
3 years ago
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to (xyz)^2 then it means that it is positive i.e f(x,y,z)\geq 0.

Consider the function F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)

We want that the gradient of this function  is equal to zero. That is (the calculations in between are omitted)

\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0

\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0

\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0

\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

-2y\lambda = -2z\lambda. If lambda is not zero, then y =z. But, since -2y\lambda=0 and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set y^2+z^2=1.

If both x,y are zero, then we have the solution set z^2=1. We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

\lambda = y^2z^2 = x^2z^2 which implies x^2=y^2

Also, from the first and third equation we have that

\lambda = y^2z^2 = x^2y^2 which implies x^2=z^2

So, in this case, replacing this in the restriction we have 3z^2=1, which gives as another solution set. On this set, we have x^2=y^2=z^2=\frac{1}{3}. Over this solution set, we have that the value of our function is \frac{1}{3^3}= \frac{1}{27}

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ale4655 [162]

Answer:

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