Weaning weights are used to evaluate di erences in growth potential of calves. A random sample of 41 steer calves is taken from a large ranch. The sample yields a mean weaning weight of 578 with a standard deviation of 87 lbs. This particular ranch is targeting an average weaning weight of 610 lbs. Perform a hypothesis test using = 0:01 to determine if the average weaning weight of cows at this ranch is di erent than 610 lbs.

Answer:

The lateral surface area of cube = 4

Step-by-step explanation:

Explanation:-

Given perimeter of the cube is 12

let 'a' be the base of cube

we know that the perimeter of cube formula = 12 a

        12 a= 12

           a = 1

The side of cube = 1cm

The lateral surface area of cube = 4 × a²

                                                      = 4 X (1)²

                                                      = 4

Final answer:-

The lateral surface area of cube = 4

Answer:

The answer is [tex]\pi(\pi-6)[/tex]

Step-by-step explanation:

Recall that green theorem is as follows: Given a field F(x,y) = (P(x,y),Q(x,y)) and a closed curve C that is counterclockwise oriented. If P and Q are continuosly differentiable, then

[tex]\oint_C F\cdot dr = \int_{R} \frac{\partial P }{\partial y}-\frac{\partial P }{\partial x} dA[/tex]

where R is the region enclosed by the curve C.

In this particular case, we have the following field [tex]F(x,y) = (4xy^2,2x^3+y)[/tex]. We are given the description of the region R as [tex]0\leq y \leq \sin(x), 0\leq x \leq \pi[/tex]. So, in this case (calculations are omitted)

[tex]\frac{\partial P}{\partial y} = 8xy, \frac{\partial Q}{\partial x} = 6x[/tex]

Thus,

[tex]\oint_C F\cdot dr =\int_{0}^{\pi}\int_{0}^{\sin(x)}(8xy-6x)dydx[/tex]

So,

[tex] \int_{0}^{\pi}\int_{0}^{\sin(x)}(8xy-6x)dydx=\int_{0}^{\pi}4x\left.y^2\right|_{0}^{\sin(x)}-6x\left.y\right|_{0}^{\sin(x)} = \int_{0}^{\pi} 4x\sin^2(x)-6x\sin(x)dx[/tex]

Since [tex]\sin^2(x) = \frac{1-\cos(2x)}{2}[/tex], then

[tex] \int_{0}^{\pi} 4x\sin^2(x)-6x\sin(x)dx = \int_{0}^{\pi} 2x(1-\cos(2x))-6x\sin(x)dx[/tex]

Consider the integrals

[tex] I_1 = \int_{0}^{\pi} x\cos(2x)dx, I_2 = \int_{0}^{\pi}x\sin(x)dx[/tex]

Then, by using integration by parts (whose calculations are omitted) we get

[tex]\int_{0}^{\pi} x\cos(2x) = \left.\frac{x\sin(2x)}{2}+\frac{\cos(2x)}{4}\right|_{0}^{\pi} = \frac{\pi\sin(2\pi)}{2}+\frac{\cos(2\pi)}{4}- \frac{0\sin(2\cdot 0)}{2}+\frac{\cos(2\cdot 0)}{4}=0[/tex]

[tex] \int_{0}^{\pi}x\sin(x) = \left.-x\cos(x)+\sen(x)\right |_{0}^{\pi} = -\pi\cos(\pi)+\sen(\pi)- (-0\cdot \cos(\pi)+\sin(0)) = \pi[/tex]

Then, we have that

[tex]\int_{0}^{\pi} 2x(1-\cos(2x))-6x\sin(x)dx = \left.x^2\right|_{0}^{\pi} -2I_1-6I_2 = \pi^2-2\cdot 0 -6\pi = \pi(\pi-6)[/tex]

Answer:

Check the explanation

Step-by-step explanation:

The fundamentals

A continuous random variable can take infinite values in the range associated function of that variable. Consider [tex]f\left( x \right)f(x)[/tex] is a function of a continuous random variable within the range [tex]\left[ {a,b} \right][a,b][/tex] , then the total probability in the range of the function is defined as:

[tex]\int\limits_a^b {f\left( x \right)dx} = 1 a∫b​ f(x)dx=1[/tex]

The probability of the function [tex]f\left( x \right)f(x)[/tex] is always greater than 0. The cumulative distribution function is defined as:

[tex]F\left( x \right) = P\left( {X \le x} \right)F(x)=P(X≤x)[/tex]

The cumulative distribution function for the random variable X has the property,

[tex]0 \le F\left( x \right) \le 10≤F(x)≤1[/tex]

The probability density function for the random variable X has the properties,

[tex]\\\begin{array}{c}\\{\rm{ }}f\left( x \right) \ge 0\\\\\int\limits_{ - \infty }^\infty {f\left( x \right)dx} = 1\\\\P\left( E \right) = \int\limits_E {f\left( x \right)dx} \\\end{array} f(x)≥0[/tex]

Kindly check the attached image below to see the full explanation to the question above.