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jenyasd209 [6]
2 years ago
14

A photocopy machine can reduce copies to 80% of their original size. By copying an already reduced copy, further reductions can

be made. (a) if a page is reduced to 80%, what percent enlargement is needed to return it to its original size? W incorrect: your answer is incorrect. Seenkey 125 % (b) estimate the number of times in succession that a page must be copied to make the final copy less than 15% of the size of the original. W incorrect: your answer is incorrect. Seenkey 9 time
Mathematics
1 answer:
alukav5142 [94]2 years ago
6 0

<u>Part a)</u> if a page is reduced to 80%, what percent enlargement is needed to return it to its original size?

Let

x---------> the percent enlargement

we know that

the original size is the 100%

so

x*80%=100%

x=(100%/80%)

x=1.25--------> 1.25=(125/100)=125%

therefore

<u>the answer Part a) is</u>

the percent enlargement is 125%

<u>Part b)</u> Estimate the number of times in succession that a page must be copied to make the final copy less than 15% of the size of the original

we know that

A photocopy machine can reduce copies to 80% of their original size

so

Copy N 1

0.80*100%=80%

Copy N 2

0.80*80%=64%

Copy N 3

0.80*64%=51.2%

Copy N 4

0.80*51.2%=40.96%

Copy N 5

0.80*40.96%=32.77%

Copy N 6

0.80*32.77%=26.21%

Copy N 7

0.80*26.21%=20.97%

Copy N 8

0.80*20.97%=16.78%

Copy N 9

0.80*16.78%=13.42%-------------> 13.42% < 15%

therefore

<u>the answer Part b) is</u>

the number of times in succession is 9


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Answer:

0.0045 = 0.45% probability that less than two of them ended in a divorce

Step-by-step explanation:

For each marriage, there are only two possible outcomes. Either it ended in divorce, or it did not. The probability of a marriage ending in divorce is independent of any other marriage. 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.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

55% of marriages in the state of California end in divorce within the first 15 years.

This means that p = 0.55

Suppose 10 marriages are randomly selected.

This means that n = 10

What is the probability that less than two of them ended in a divorce?

This is

P(X < 2) = P(X = 0) + P(X = 1)

In which

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

P(X = 0) = C_{10,0}.(0.55)^{0}.(0.45)^{10} = 0.0003

P(X = 1) = C_{10,1}.(0.55)^{1}.(0.45)^{9} = 0.0042

P(X < 2) = P(X = 0) + P(X = 1) = 0.0003 + 0.0042 = 0.0045

0.0045 = 0.45% probability that less than two of them ended in a divorce

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1 year ago
A normal distribution curve, where x = 70 and σ = 15, was created by a teacher using her students’ grades. What information abou
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Answer:

The median and mode of the students grade is 70.

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From the provided information it can be seen that the mean of the distribution is, <em>μ</em> = 70 and the standard deviation is, <em>σ</em> = 15.

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So, the median and mode of the students grade is 70.

The standard deviation of the data represents the spread of the observation, i.e. how dispersed the values are along the curve.

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P(\mu-2\sigma

P(\mu-3\sigma

Assuming that maximum marks of the exam is 100, it can be said that most of the students scored between 40 and 100.

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A spherical scoop of ice cream is placed on top of a hollow ice cream cone. the scoop and cone have the same radius. the ice cre
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The figure shown below illustrates the problem.

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V₁ = (1/3) π r²h

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V₂ = (4/3) π r³

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V₁ = V₂
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What is the multiplicative inverse of -0.7?
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The parallel dotplots below display the girths (belly
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Answer:

The standard deviation for the distribution of girths is

about the same for both male and female pigs.

Step-by-step explanation:

Step 1

We will interprete the dotplots

For the Male pigs

24, 26, 26, 26, 26, 26, 26, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 30, 30, 30, 30, 30, 30, 32

Range

Maximum value - Minimum value

= 32 - 24

= 8

IQR = Interquartile range = Q3 - Q1

Q1 formula = 1/4(n + 1)th

=n = 25

= 1/4(25 + 1)

= 1/4(26)

= 26/4

= 6.5

This mean the value is between the 6th and 7th value

6th value = 26

7th value = 26

Q1 = 26 + 26/2

Q1 = 26

Q3

Q3 formula = 3/4(n + 1)th

=n = 25

= 3/4(25 + 1)

= 3/4(26)

= 78/4

= 19.5

This mean the value is between the 19th and 20th value

19th value = 30

20th value = 30

Q3= 30+ 30/2

= 30

IQR = 30 - 26

= 4

Standard deviation

Mean

First we find the mean

= 24+ 26+26+ 26+ 26+ 26+26+28+ 28+ 28+ 28+28+ 28+ 28+28+ 28+28+28+30+ 30+30+ 30+30+30+32/28

= 700/25

= 28

Standard deviation Formula = √(x - mean)²/n - 1

= √[(24-28)² + (26- 28)² + (26 -28)² + (26 - 28)² +................ (32 - 28)²]/25 - 1

= √(16 + 4 + 4+ 4+ 4+4+ 4+0+ 0+ 0+ 0+0+ 0+ 0+ 0+0+0+0+ 4+ 4+4+4+4+ 4+ 16)/25 - 1

= √80/25 - 1

=√3.333333333

1.825741858

Female Pigs

21, 23, 23, 23,23,23,23,25, 25, 25, 25, 25, 25, 25, 25, 25, 25,25, 27, 27, 27, 27,27, 27, 29

Range:

Maximum value - Minimum value

= 29 - 21

8

IQR

IQR = Interquartile range = Q3 - Q1

Q1 formula = 1/4(n + 1)th

=n = 25

= 1/4(25 + 1)

= 1/4(26)

= 26/4

= 6.5

This mean the value is between the 6th and 7th value

6th value = 23

7th value = 23

Q1 = 23 + 23/2

Q1 = 23

Q3

Q3 formula = 3/4(n + 1)th

=n = 25

= 3/4(25 + 1)

= 3/4(26)

= 78/4

= 19.5

This mean the value is between the 19th and 20th value

19th value = 27

20th value = 27

Q3= 27+ 27/2

Q3 = 27

IQR = 27 - 23

= 4

First we find the Mean

=(21+23+23+23+23+23+23+25+ 25+ 25+ 25+25+ 25+ 25+ 25+25+25+25+27+27+ 27+27+27+ 27+29)/25

= 625/25

= 25

Standard deviation Formula = √(x - mean)²/n - 1

= √[(21-25)² + (23- 25)² + (23 -25)² + (23 - 25)² +................ (29 - 25)²]/25 - 1

= √(16 + 4 + 4+ 4+ 4+4+ 4+0+ 0+ 0+ 0+0+ 0+ 0+ 0+0+0+0+ 4+ 4+4+4+4+ 4+ 16)/25 - 1

= √80/25 - 1

=√3.333333333

1.825741858

Looking at the calculation above and comparing both the girths from the male and female

We can conclude that: The standard deviation for the distribution of girths is

about the same for both male and female pigs which is 1.825741858

5 0
2 years ago
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