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Volgvan
3 years ago
15

You will begin with a relatively standard calculation. consider a concave spherical mirror with a radius of curvature equal to 6

0.0 centimeters. an object 6.00 centimeters tall is
Mathematics
1 answer:
vodomira [7]3 years ago
6 0
We are given a concave spherical mirror with the following dimensions:

Radius = 60 cm; D o = 30 cm
Height = 6 cm; h o = 6 cm

First, we need to know the focal length, f, of the object (this should be given). Then we can use the following formulas for calculation: 

Assume f = 10 cm

1/ f = 1 /d o + 1 / d i
1 / 10 = 1 / 30 + 1 / d i

d i = 15 cm

Then, calculate for h i:

h i / h o = - d i / d o

h i / 6 = - 15 / 30

h i = - 3 cm

Therefore, the distance of the object from the mirror is 3 centimeters. The negative sign means it is "inverted". 
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seraphim [82]

Answer:

\text{D) }99^{\circ}

Step-by-step explanation:

Define a cyclic quadrilateral by a quadrilateral that is circumscribed by a circle. In this case, since the quadrilateral shown is circumscribed by a circle, it is a cyclic quadrilateral.

A property of all cyclic quadrilaterals is that their opposite angles are supplementary, meaning they add up to 180 degrees. Since \angle G and \angle I are opposite angles in the quadrilateral, they must be supplementary. Therefore, we have the equation:

\angle G+\angle I=180,\\81^{\circ}+\angle I=180,\\\angle I=180-81=\boxed{99^{\circ}}

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3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample
omeli [17]

Given Information:

Population mean = p  = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

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Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10  (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

\sigma = \sqrt{\frac{p(1-p)}{n} }

Where p is the population mean that is proportion of female students and n is the sample size.

\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma =  0.069

Therefore, the standard deviation of the sampling distribution is 0.069.

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What is 5-5menemensnsn
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Answer:

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