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Licemer1 [7]
2 years ago
12

The magnification produced by a plane mirror is +1. what does this mean?

Physics
1 answer:
disa [49]2 years ago
6 0

Answer:

The magnification produced by a plane mirror is +1

means then the size of the image is equal to the size of the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object.

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Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl
ollegr [7]
<h2>Answer:</h2>

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

<h2>Explanation:</h2><h2 />

<em>Given frequency table (find attached as Table 1);</em>

<u></u>

(a) To find the sample standard deviation and sample variance, follow these steps;

<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

<em>So the new table becomes (find attached as Table 2);</em>

<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>

n = 50+18+42+20+46

n = 176

<em>iii. Calculate the mean (M)</em>

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

<em>iv. Find the variance (σ²);</em>

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)                ------------(i)

Where;

f = frequency of each boundary data point

<em>=>  Let's first calculate </em>Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

<em>=> Now calculate (n x M²)</em>

n x M² = 176 x 11.44²

n x M² = 23033.7536

<em>=> Now substitute these values into equation (i) to calculate the variance</em>

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

<em>v. Find the standard deviation (σ)</em>

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

4 0
2 years ago
Conductors have ___<br> resistance.
vazorg [7]

Answer:

little/no

Explanation:

Conductors are materials, which conduct electricity and/or heat. That means, that their resistance to such energy is so little, that an electric current is able to pass through.

5 0
3 years ago
What is the acceleration of a baseball that has a final velocity of 15.0 m/s when it crosses the plate 2.0 seconds after leaving
Leokris [45]
I think it’s A pretty sure...
3 0
3 years ago
Which of the following statements apply to electric charges?
Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0
3 years ago
A very long straight wire has charge per unit length 1.44×10-10C/m.
4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m            

Explanation:

We have given charge per unit length, that is liner charge density \lambda =1.44\times 10^{-10}C/m

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

E=\frac{\lambda }{2\pi \epsilon _0r}, here \lambda is linear charge density and r is distance of the point where we have to find the electric field

So 2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0
3 years ago
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