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torisob [31]
2 years ago
10

The ratio of Michaels's points on an English test to Paloma's points on the same test is 12:11. The ratio of Michaels's points o

n the English test on the same test was 4:3. the 3 students score a total of 224 points on the same test. What is the difference between Palomas and Antons' points on the English test? Try to explain fully, if possible. Thank you SO MUCH!
Mathematics
1 answer:
diamong [38]2 years ago
3 0

The difference between Palomas and Antons' points on the English test is of 14 points.

<h3>What is Ratio ?</h3>

When two numbers can be written in the form as p:q , they are said to be in Ratio.

It is given that

The ratio of Michaels point to Paloma point is 12 :11

The ratio of Michaels point to Anton's Point is 4:3

Let Paloma's point be x

then Michaels point will be 12x/11

and Anton's points will be

(12x/11 ) / Anton's points = 4 /3

Anton's points = 9x/11

It is given that the sum of the points of the three students is 224

so

x + (12x/11)+ (9x/11) = 224

32x /11 = 224

x = 77

Paloma's points = 77

Anton's Points = 9x/11 = 63

The difference between Palomas and Antons' points on the English test

=77 -63

= 14 points.

To know more about Ratio

brainly.com/question/13419413

#SPJ1

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The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a s
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a) P(X\geq 1) = 1-P(X

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And replacing we got:

P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71

Step-by-step explanation:

For this case we can solve this problem creating the following table

Number of particles      Frequency         Rel. Frequency

             0                            1                       1/100 =0.01

             1                             2                      2/100 =0.02

             2                            3                       3/100=0.03

             3                            12                     12/100=0.12    

             4                            11                       11/100=0.11

             5                            15                      15/100=0.15

             6                            18                      18/100=0.18

             7                            10                       10/100=0.1

             8                            12                       12/100=0.12

             9                            4                         4/100=0.04

             10                           5                         5/100=0.05

             11                            3                         3/100=0.03

             12                           1                          1/100=0.01

             13                           2                         2/100=0.02

             14                           1                          1/100=0.01

           Total                       100                            1

We assume on this case the the relative frequency represent the probability.

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For this case we can use the complement rule like this:

P(X\geq 1) = 1-P(X

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Again for this case we can use the complement rule and we got:

P(X\geq 5) = 1-P(X

And replacing we got:

P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71

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