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attashe74 [19]
2 years ago
13

20 POINTS

Mathematics
1 answer:
TiliK225 [7]2 years ago
3 0

Answer:

Step-by-step explanation:

There are <u>6</u><u> different shapes</u>

You want the outcome to be a Nonagon

You put the outcome as a ratio 1/6

1/6=0.1666667

0.1666667*100=16.6667%

<u>Chance of pulling out a </u><u>nonagon</u>

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Circle J is located in the first quadrant with center (a, b) and radius s. Felipe transforms Circle J to prove that it is simila
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The correct answer for the question that is being presented above is this one: "c. translate circle J by (x-a, y-b) and dilate by a factor of t/s." Circle J is located in the first quadrant with center (a, b) and radius s. Felipe transforms Circle J to prove that it is similar to any circle centered <span>at the origin with radius r.</span>
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2 years ago
A worker has a base pay of $20 per hour, gets $3 more per hour after 8 hours, and gets $4 more
GalinKa [24]

Answer:

177 $

Step-by-step explanation:

20 x8 = 160

3x3= 9

4x2=8

8+9+160= 177

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2 years ago
What is the equation of the line? Iready
leva [86]

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C

Step-by-step explanation:

The y intercept is 2 and if you do the rise/run you'll find that the slope has to be 1/2

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1/2x+2

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1 year ago
Read 2 more answers
The quality control team of a company checked 800 digital cameras for defects. The team found that 20 cameras had lens defects,
dmitriy555 [2]

Answer:

0.025

Step-by-step explanation:

-This is a conditional probability problem.

-Let L denote lens defect and C charging defect.

#We first calculate the probability of a camera having a lens defect;

P(lens)=\frac{Lens}{Total}\\\\=\frac{20}{800}\\\\=0.025

#Calculate the probability of a camera having a charging defect:

P(Charging)=\frac{Charging}{Total}\\\\=\frac{25}{800}\\\\=0.03125

The  the probability that a camera has a lens defect given that it has a charging defect is calculated as:

P(L|C)=\frac{P(C)P(L)}{P(C)}\\\\=\frac{0.025\times 0.03125}{0.03125}\\\\=0.025

Hence,  the probability that a camera has a lens defect given that it has a charging defect is 0.025

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2 years ago
The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
Marina86 [1]

Answer:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

3 0
2 years ago
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