Write and solve an inequality to find the possible values of x.

Using Maxwell's equations, we can determine the magnetic flux density. One of the Maxwell's equations is:

[tex]\displaystyle \nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}[/tex],

where [tex]\displaystyle \nabla \times \mathbf{H}[/tex] is the curl of the magnetic field intensity [tex]\displaystyle \mathbf{H}[/tex], [tex]\displaystyle \mathbf{J}[/tex] is the current density, and [tex]\displaystyle \frac{\partial \mathbf{D}}{\partial t}[/tex] is the time derivative of the electric displacement [tex]\displaystyle \mathbf{D}[/tex].

In this problem, there is no current density ([tex]\displaystyle \mathbf{J} =0[/tex]) and no time-varying electric displacement ([tex]\displaystyle \frac{\partial \mathbf{D}}{\partial t} =0[/tex]). Therefore, the equation simplifies to:

[tex]\displaystyle \nabla \times \mathbf{H} =0[/tex].

Taking the curl of the given magnetic field intensity [tex]\displaystyle \mathbf{R} =2\cos( 10^{10} t-600x)\hat{a}_{z}\, \text{Am}[/tex]:

[tex]\displaystyle \nabla \times \mathbf{R} =\nabla \times ( 2\cos( 10^{10} t-600x)\hat{a}_{z}) \, \text{Am}[/tex].

Using the curl identity and applying the chain rule, we can expand the expression:

[tex]\displaystyle \nabla \times \mathbf{R} =\left( \frac{\partial ( 2\cos( 10^{10} t-600x)) \hat{a}_{z}}{\partial y} -\frac{\partial ( 2\cos( 10^{10} t-600x)) \hat{a}_{z}}{\partial z}\right) \mathrm{d} x\mathrm{d} y\mathrm{d} z[/tex].

Since the magnetic field intensity [tex]\displaystyle \mathbf{R}[/tex] is not dependent on [tex]\displaystyle y[/tex] or [tex]\displaystyle z[/tex], the partial derivatives with respect to [tex]\displaystyle y[/tex] and [tex]\displaystyle z[/tex] are zero. Therefore, the expression further simplifies to:

[tex]\displaystyle \nabla \times \mathbf{R} =-\frac{\partial ( 2\cos( 10^{10} t-600x)) \hat{a}_{z}}{\partial x} \mathrm{d} x\mathrm{d} y\mathrm{d} z[/tex].

Differentiating the cosine function with respect to [tex]\displaystyle x[/tex]:

[tex]\displaystyle \nabla \times \mathbf{R} =-2( 10^{10}) \sin( 10^{10} t-600x)\hat{a}_{z} \mathrm{d} x\mathrm{d} y\mathrm{d} z[/tex].

Setting this expression equal to zero according to [tex]\displaystyle \nabla \times \mathbf{H} =0[/tex]:

[tex]\displaystyle -2( 10^{10}) \sin( 10^{10} t-600x)\hat{a}_{z} \mathrm{d} x\mathrm{d} y\mathrm{d} z =0[/tex].

Since the equation should hold for any arbitrary values of [tex]\displaystyle \mathrm{d} x[/tex], [tex]\displaystyle \mathrm{d} y[/tex], and [tex]\displaystyle \mathrm{d} z[/tex], we can equate the coefficient of each term to zero:

[tex]\displaystyle -2( 10^{10}) \sin( 10^{10} t-600x) =0[/tex].

Simplifying the equation:

[tex]\displaystyle \sin( 10^{10} t-600x) =0[/tex].

The sine function is equal to zero at certain values of [tex]\displaystyle ( 10^{10} t-600x) [/tex]:

[tex]\displaystyle 10^{10} t-600x =n\pi[/tex],

where [tex]\displaystyle n[/tex] is an integer. Rearranging the equation:

[tex]\displaystyle x =\frac{ 10^{10} t-n\pi }{600}[/tex].

The equation provides a relationship between [tex]\displaystyle x[/tex] and [tex]\displaystyle t[/tex], indicating that the magnetic field intensity is constant along lines of constant [tex]\displaystyle x[/tex] and [tex]\displaystyle t[/tex]. Therefore, the magnetic field intensity is uniform in the given medium.

Since the magnetic flux density [tex]\displaystyle B[/tex] is related to the magnetic field intensity [tex]\displaystyle H[/tex] through the equation [tex]\displaystyle B =\mu H[/tex], where [tex]\displaystyle \mu[/tex] is the permeability of the medium, we can conclude that the magnetic flux density is also uniform in the medium.

Thus, the correct expression for the magnetic flux density in the given medium is:

[tex]\displaystyle B =6\cos( 10^{10} t-600x)\hat{a}_{z}[/tex].

84.13% of the printing jobs will be acceptable.

The new standard deviation required to achieve a minimum of 95% of jobs acceptable is 0.121.

The darkness of the print is measured quantitatively using an index. If the index is greater than or equal to 2.0 then the darkness is acceptable. Anything less than 2.0 means the print is too light and not acceptable. The machines print at an average darkness of 2.2 with a standard deviation of 0.20.

The mean of the darkness of the print is µ = 2.2 and the standard deviation is σ = 0.20.Therefore, the z-score can be calculated as; `z = (x - µ) / σ`.The index required for acceptable prints is 2.0. Thus, the percentage of prints that are acceptable can be calculated as follows;P(X ≥ 2.0) = P((X - µ)/σ ≥ (2.0 - 2.2) / 0.20)P(Z ≥ -1) = 1 - P(Z < -1)Using the standard normal table, P(Z < -1) = 0.1587P(Z ≥ -1) = 1 - 0.1587= 0.8413.

To find the new standard deviation, we can use the z-score formula.z = (x - µ) / σz = (2.0 - 2.2) / σz = -1Therefore, P(X ≥ 2.0) = 0.95P(Z ≥ -1) = 0.95P(Z < -1) = 0.05Using the standard normal table, the z-score value of -1.645 corresponds to a cumulative probability of 0.05. Hence,z = (2.0 - 2.2) / σ = -1.645σ = (2.0 - 2.2) / -1.645= 0.121.

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