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1 year ago

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An object of height 8.50 cm is placed 20.0 cm to the left of a converging lens with a focal length of 12.0 cm. Determine the ima

ge location in cm, the magnification, and the image height in cm.

1 answer:

ddd [48]1 year ago

5 0

The image distance is 33.3 cm while the image height is 14.2 cm.

<h3>What is a converging lens?</h3>

A converging lens will always have a positive focal length hence, we have to find the object distance as follows;

1/f = 1/v + 1/u

1/12 = 1/v + 1/20

1/v = 1/12 - 1/20

1/v = 0.08 - 0.05

v =33.3 cm

Now;

Magnification = 33.3 cm/20.0 cm =1.67

M = Image height/Object height

1.67 = Image height/8.50 cm

Image height = 1.67 * 8.50 cm

Image height = 14.2 cm

Learn more about focal length:brainly.com/question/16188698

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Answer:

a) $P(6)=0.25$

b) $p(x$

c) $p(x\geq3)=0.9878$

d) $\sigma=\sqrt{2.4}=1.5492$

Explanation:

From the question we are told that:

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Sample size $n=10$

Let x =customers ask for water

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Generally the equation for y is mathematically given by

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Generally the equation for pmf p(x) is mathematically given by

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a)

Generally the probability that exactly 6 ask for water is mathematically given by

$P(x)6=10C_6 (0..6)^6(0.4)^{10-6}$

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b)

Generally the probability that less than 9 ask for water with meal is mathematically given by

$p(xg)$

$p(x$

$p(x$

$p(x$

c)

Generally the probability that at least 3 ask for water with meal is mathematically given by

$p(x\geq3)=1-p(x$

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Generally the mean and standard deviation of sample size is mathematically given by

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Standard deviation

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Explanation:

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Let $v$ be the velocity of fluid at a point.

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