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zhuklara [117]
3 years ago
5

The total charge on 6 particles is -48units. All the particles have the same charge. What is the charge on each particle?

Mathematics
1 answer:
Marina CMI [18]3 years ago
4 0

i believe the answer will be -7

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How to find the average squared distance between the points of the unit disk and the point (1,1)
ddd [48]
The unit disk can be parameterized by the function

\mathbf p(r,\theta)=(r\cos\theta,r\sin\theta)

where 0\le r\le 1 and 0\le\theta\le2\pi. The squared distance between any point in this region (x,y)=(r\cos\theta,r\sin\theta) and the point (1, 1) is

(x-1)^2+(y-1)^2=(r\cos\theta-1)^2+(r\sin\theta-1)^2
=(r^2\cos^2\theta-2r\cos\theta+1)+(r^2\sin^2\theta-2r\sin\theta+1)
=r^2(\cos^2\theta+\sin^2\theta)-2r(\cos\theta-\sin\theta)+2
=r^2-2r(\cos\theta-\sin\theta)+2

The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.

\dfrac{\displaystyle\iint_{x^2+y^2
=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}
=\dfrac{\frac{5\pi}2}\pi=\dfrac52
5 0
3 years ago
A gardener wants to know if soaking seeds in water before planting them increases the proportion of seeds that germinate. To inv
juin [17]

Answer:

d. I and III only

Step-by-step explanation:

I. The seeds should be randomly assigned to a treatment.

III. The number of successful seeds and unsuccessful seeds in each group should be at least 10.

The distribution of difference between two sample proportions :

Given :

Proportion 1 = P1 ;

Proportion 2 = P2 ;

Sample assignment for both samples 1 and 2 into the different treatment groups should be randomized, that is a simple random sampling of subjects into the treatment and control group. The sample design for difference between two sample proportions should be independent.

Finally each of the two proportions P1 and P2 should record a minimum of 10 successes and 10 non - successful Occurrences.

3 0
3 years ago
how many such tests would it take for the probability of committing at least one type i error to be at least 0.7? (round your an
drek231 [11]

The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118   .

In the question ,

it is given that ,

the probability of committing at least , type I error is = 0.7

we  have to find the number of tests ,

let the number of test be n ,

the above mentioned situation can be written as

1 - P(no type I error is committed) ≥ P(at least type I error is committed)

which is written as ,

1 - (1 - 0.01)ⁿ ≥ 0.7

-(0.99)ⁿ ≥ 0.7 - 1

(0.99)ⁿ ≤ 0.3

On further simplification ,

we get ,

n ≈ 118 .

Therefore , the number of tests are 118 .

Learn more about Error here

brainly.com/question/17062640

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4 0
1 year ago
There are 812,000 gallons of water in an Olympic training center’s pool. Express this number in scientific notation
DerKrebs [107]
8.12 ×10^5 is the answer
7 0
3 years ago
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Mitchell is measured in February for the customs of a school play he’s 47 feet fabric he has 12 1/3 yards of fabric how many mor
Step2247 [10]
47- 12 1/3 = 34 2/3
He needs 34 and 2/3 feet more of fabric
8 0
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