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Iteru [2.4K]
3 years ago
6

In interval notation signs use what grouping symbol?

Mathematics
1 answer:
Vesnalui [34]3 years ago
3 0

Answer:

either parentheses or brackets

Step-by-step explanation:

(x,y) means that the number is not included in interval notation

[x,y] means that the number is included in interval notation

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Find the​ P-value for the indicated hypothesis test with the given standardized test​ statistic, z. Decide whether to reject H0
Leviafan [203]

Answer: We do not reject the null hypothesis.

Step-by-step explanation:

  • When the p-value is greater than the significance level , then we do not reject the null hypothesis or if p-value is smaller than the significance level , then we reject the null hypothesis.

Given : Test statistic : z = -2.28

Significance level : \alpha=0.02

By using the standard normal distribution table ,

The p-value corresponds to the given test statistic ( two tailed ):-

2P(Z>|z|)=2P(Z>2.28)=0.0226

Since the p-value is greater than the significance level of 0.02.

Then , we do not reject the null hypothesis.

8 0
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
PLEASEEEE HELPPPP<br><br><br>Add. 3+(-7)=<br>​
kow [346]

Answer:

Step-by-step explanation:

It’s exactly like saying -7 + 3.. -7 is the bigger number so you take away from that the answer is -4

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3 years ago
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Geometry !!! Pls help it’s timed!!
DENIUS [597]

Answer:

the answer is the first choice .

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2 years ago
For the function f(x) = x + 7, what is the ordered pair for the point on the graph when x = 2b. A. ( 2b, 2b+7). B. ( 2b, x + 7).
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f(2b)=2b+7\\\\&#10;(2b,2b+7)
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