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Vsevolod [243]
3 years ago
14

D(n) = 8 - 6(n - 1)d(n)=8−6(n−1)

Mathematics
1 answer:
VikaD [51]3 years ago
8 0

Answer: 2

Step-by-step explanation:

So you distribute the -6 to the n and -1

Resulting info d(n)= 8-6n+6

Combine the 8 and 6. Getting 14

Add 6n in both sides getting

7n=14

Divide by 7

N=2

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Apple weights in an orchard are normally distributed. from a sample farmer fred determines the mean weight of a box of apples to
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<span>Mean = 270 Standard deviation = 10 x = 255 Formula for z-score, z = (x - mean)/SD z = (255 - 270) / 10 => z = -15 / 10 => z = -1.5 So by referring to z-table, -1.5 correlates to 0.0668 that implies to 0.07 So 7% of the boxes of Apples weight less than 255oz. The percentage of boxes is in the range of 255 oz and 270 oz, Now calculating the requiring percentage 50% - 7% = 43%</span>
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3 years ago
A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and
natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: E=k\frac{q}{r^{2} }; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: p=\frac{q}{A}; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: S=4\pi r^{2}; which gives us the surface of a sphere.

The inner surface: Si=4\pi (0.06m)^{2} = 0.04 m^{2}

The outer surface: S=4\pi (0.08m)^{2}=0.08m^{2}

Now we can calculate the charges,

Inner charge: Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC

Outer charge: Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC

Then, we are able to calculate both fields:

Inner field: Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}

Outer field:  Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0
3 years ago
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