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Harrizon [31]
3 years ago
6

I need the answer so I can pass the first semester‍♀️

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer:

Step-by-step explanation:

0.333333 = 1/3

can check through dividing 1 by 3

0.125 = 125/1000 = 25/200 = 5/40 = 1/8

0.166666 = 1/6

0.1 = 1/10

0.66666 = 2/3

0.2 = 2/10 = 1/5

0.75 = 75/100 = 15/20 = 3/4

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The line segment joining A(6, 3) to B(–1, –4) is doubled in length by having half its length added to each end. Find the coordin
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Answer:

  • (9.5, 6.5) and (-4.5, -7.5)

Step-by-step explanation:

Let the extended points be A' and B' and add the point M as midpoint of AB

<u>Coordinates of M are:</u>

  • ((6 - 1)/2, (3-4)/2) = (2.5, -0.5)

Now point A is midpoint of A'M and point B is midpoint of MB'

<u>Finding the coordinates using midpoint formula:</u>

  • A' = ((2*6 - 2.5),(2*3 - (-0.5)) = (9.5, 6.5)
  • B' = ((2*(-1) - 2.5), (2*(-4) - (-0.5)) = (-4.5, -7.5)

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What is the difference of 23x and 9x?
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There are 10 true-false questions and 20 multiple choice questions from which to choose a five-question quiz how many ways can t
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Answer:

In 68229 ways can the quiz be selected such that there is atleast three multiple choice questions

Step-by-step explanation:

Given:

Number of True or false questions= 10

Number of multiple choice questions= 20

To Find:

How many ways can 5 questions can be selected if there must be at least three multiple choice questions =?

Solution:

Combination

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.  

The question States there sholud be ATLEAST 3 multiple choice question,

So, we may have

(3 Multiple choice question and 2 true or false question) or

(4 Multiple choice question and 1 true or false question) or

(5 Multiple choice question and 0 true or false question)

Required Number of ways = (20C3 X10C2) +(20C4 X10C1) + (20C5 X10C0)

Required Number of ways =(\frac{20!}{20!(20-3)!}\times\frac{10!}{10!(10-2)!})+(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-1)!}) +(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-0)!})

Required Number of ways = ( 1140 x 42) + (4845 x 10) +(15504 x 1)

Required Number of ways = 47880+48450+15504

Required Number of ways = 68229

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3 years ago
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