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san4es73 [151]
3 years ago
6

A microscope has a setting that magnifies an object so that it appears 100 times as large when veiwed through the eyepiece. If a

dust particle is 0.03 cm long, how long will the dust particle appear in cenimeters through the microscope? what about 1000 times as large
Mathematics
1 answer:
postnew [5]3 years ago
6 0
100 times as large would be 3 cm and 100 times as large would be 30 cm
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Answer:

7/8.

Step-by-step explanation:

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Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

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p = \frac{31}{365} = 0.0849

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This means that n = 20

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This is P(X = 2).

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0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

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3 years ago
I will send a pic to you​
Citrus2011 [14]

Answer:

agreed

Step-by-step explanation:

agreed...............................

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Answer:

Here's your solution

=> <em>Both </em><em>angles</em><em> </em><em>are </em><em>interior</em><em> </em><em>angles</em><em> </em><em>hence</em><em> </em><em>their</em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>sum </em><em>will </em><em>equal</em><em> to</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em> </em><em>(</em><em>2</em><em>x</em><em> </em><em>+</em><em> </em><em>1</em><em>0</em><em>)</em><em>°</em><em> </em><em> </em><em>+</em><em> </em><em>(</em><em> </em><em>x </em><em>+</em><em> </em><em>5</em><em>)</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em>

<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em> </em><em>2</em><em>x</em><em> </em><em>+</em><em> </em><em>1</em><em>0</em><em>°</em><em> </em><em>+</em><em> </em><em>x </em><em>+</em><em> </em><em>5</em><em>°</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em>

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<em> </em><em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em> </em><em>3</em><em>x</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em><em> </em><em>-</em><em> </em><em>1</em><em>5</em><em>°</em>

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<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em>. </em><em>x </em><em>=</em><em> </em><em>1</em><em>6</em><em>5</em><em>/</em><em>3</em>

<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em>. </em><em>x </em><em>=</em><em> </em><em>5</em><em>5</em><em>°</em>

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