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fgiga [73]
2 years ago
12

N a certain region, the electric potential due to a charge distribution is given by the equation v(x,y,z)=3(x^2)(y^2) +y(z^3) -

2x(z^3), where x,y, and z are measured in meters and v is in volts. calculate the magnitude of the electric field vector at the position (x,y,z)=(1.0,1.0,1.0).
Mathematics
1 answer:
katrin [286]2 years ago
6 0

The electric field \mathbf E is the negative of the gradient of the electric potential:

\mathbf E=-\nabla V(x,y,z)=-\left(6xy^2-2z^3,6x^2y+z^3,3yz^2-6xz^2\right)

At the point (1.0, 1.0, 1.0), the electric field has magnitude

\|\mathbf E\|=\|(4.0,7.0,-3.0)\|=8.6\dfrac{\rm V}{\rm m}

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A scientist measured a museum specimen. She recorded the measurement as 75 centimeters. What could the scientist have been measu
saw5 [17]

Answer:

75cm is the measurment

Step-by-step explanation:

It would have had to have been the lenght of the gem because 75cm or 75 centimeters is a measurment not the mass of something. The mass is the amount of matter in an object.

8 0
2 years ago
The sum of Sasha’s and Natalie’s ages is 31. Twice Natalie’s age is 11 years more than Sasha’s age. Find each of their ages
ZanzabumX [31]

Answer:

Natalie's age is 14

Sasha's age is 17

Step-by-step explanation:

Let

Sasha's age be "s", and

Natalie's age be "n"

<u>Sum of Sasha and Natalie is 31:</u>

s + n = 31

<u>Twice Natalie is 11 more than Sasha:</u>

2n = s + 11

We ca solve this equation for s first:

s = 2n - 11

We put this into first equation and solve for n first:

s + n = 31

2n - 11 + n = 31

3n = 31 + 11

3n = 42

n = 42/3 = 14

Natalie's age is 14

Also, s = 2n - 11

s = 2(14) - 11

s = 28 - 11

s = 17

Sasha's age is 17

6 0
3 years ago
The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament.
NeX [460]

Answer:

Step-by-step explanation:

Hello!

(Data and full text attached)

The stem and leaf plot is a way to present quantitative data.  Considering two-digit numbers, for example 50, the tens digits are arranged in the stem and the units determine the leafs.

So for the stem and leaf showing the digs of the top players of the tournament, the observed data is:

41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97

n= 15

Note that in the stem it shows the number 8, but with no leaf in that row, that means that there were no "eighties" observed.

a) 6 Players had more than 60 digs.

b)

To calculate the mean you have to use the following formula:

X[bar]= ∑x/n= (41 + 41 + 43 + 43 + 45 + 50 + 42 + 53 + 54 + 62 + 63 + 63 + 67 + 75 + 97)/15= 849/15= 56.6 digs

To calculate the median you have to calculate its position and then identify its value out of the observed data arranged from least to greatest:

PosMe= (n+1)/2= (15+1)/2= 8 ⇒ The median is in the eight place:

41, 41, 43, 43, 45, 50, 42, 53, 54, 62, 63, 63, 67, 75, 97

The median is Me= 53

53 is the value that separates the data in exact halves.

The mode is the most observed value (with more absolute frequency).

Consider the values that were recorded more than once

41, 41

43, 43

63, 63

41, 43 and 63 are the values with most absolute frequency, which means that this distribution is multimodal and has three modes:

Md₁: 41

Md₂: 43

Md₃: 63

The Range is the difference between the maximum value and the minimum value of the data set:

R= max- min= 97 - 41= 56

c)

The distribution is asymmetrical, right skewed and tri-modal.

Md₁: 41 < Md₂: 43 < Me= 53 < X[bar]= 56.6 < Md₃: 63

Outlier: 97

d)

An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.  

Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:

X < Q₁ - 1.5 IQR

X > Q₃ + 1.5 IQR

PosQ₁= 16/4= 4

Q₁= 43

PosQ₃= 16*3/4= 12

Q₃= 63

IQR= 63 - 43= 20

Q₁ - 1.5 IQR = 43 - 1.5*20= 13 ⇒ There are no values 13 and below, there are no lower outliers.

Q₃ + 1.5 IQR = 63 + 1.5*20= 93 ⇒ There is one value registered above the calculated limit, the last observation 97 is the only outlier of the sample.

The mean is highly affected by outliers, its value is always modified by the magnitude of the outliers and "moves" its position towards the direction of them.

Calculated mean with the outlier: X[bar]= 849/15= 56.6 digs

Calculated mean without the outlier: X[bar]= 752/14= 53.71 digs

I hope this helps.

7 0
3 years ago
What’s pi? (i actually kno what it is i’m jus bored lol)
Alenkasestr [34]

Answer:

3.14159 and so on

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Bella and Heather put some money into their money boxes every week. The amount of money (y), in dollars, in their money boxes af
Hatshy [7]

Answer:

\boxed{ \text{B. 10 weeks, \$310 }}\\

Step-by-step explanation:

We have two conditions:

\begin{array}{lrcll}(1) & y & = & 25x + 60 & \\(2) & y & = & 30x + 10 & \\ & 25x + 60 & = & 30 x + 10 & \text {Set (1) = (2)} \\ & 50 & = & 5x & \text{Subtracted 25x from each side} \\& x & = & 10 &\text{Divided each side by 5} \\\end{array}\\$ $\\\boxed{ \textbf{The number of weeks is }x = 10}\\

   

    Bella: y = 25x + 60 = 25(10) + 60 = 250 + 60 = 310  

Heather: y = 30x + 10  = 30(10) +  10 = 300 + 10  = 310

7 0
3 years ago
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