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postnew [5]
3 years ago
10

Please Help ME!!!

Mathematics
1 answer:
Airida [17]3 years ago
3 0

Answer:

Hence Group A compose of the alternate Addition of  "ONE" and "THREE" units

Hence Group B Composes of alternate 22 units addition and 3 units addition.

Step-by-step explanation:

Given:

Two groups as A and B with number in certain patterns as

To Find:

What is that certain pattern in numbers?

Solution:

Group A includes the following numbers as ,

0,1,4,5,8

Here 1st number=0 then 2nd number =1

So in between this addition is 1 is present

And then 3rd number =4 and 4th number=5

so there  alternate addition of "one"(1) present

Now in 3rd and 2nd number there is addition of 3 units

And in 5th and 4 number there is addition of 3 units .

Hence this group compose of the alternate Addition of  "ONE" and "THREE" units.

Group B includes the following number as ,

10,32,35,57,79

So here 1st number is 10 and 2nd number is 32 (Addition of 22 units).

Further alternate 4th 57 and 3rd number is 35 (addition of 22 units).

For 3rd number is 35 and 2nd number is 32 (addition of 3 units)

For<em> 6th number will be is  82 </em>and  5th number is 79 (addition of 3 units).

Hence this group Composes of alternate 22 units addition and 3 units addition.

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Step-by-step explanation

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3 years ago
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alina1380 [7]
For this case we can make use of the Pythagorean theorem.
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 Rewriting:
 x = root (117-36)
 x = root (81)
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4 0
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Ghella [55]
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5 0
3 years ago
Exercise 1.A committee of five people is chosen randomly from four men and six women. Find the probability that:a) Exactly four
MatroZZZ [7]

Answer with explanation:

Given : Number of men = 4

Number of women = 6

Total people = 6+4=10

Total number of ways to make a committee of five people from 10 persons :-

^{10}C_{4}=\dfrac{10!}{4!(10-4)!}=210

a) Number of ways to make committee that has exactly four women :

^6C_4\times ^4C_1=\dfrac{6!}{4!(6-4)!}\times4=60

The  probability that committee has exactly four women :

\dfrac{60}{210}=\dfrac{2}{7}

b) Number of ways to make committee that has at-least four women :

^6C_4\times ^4C_1+^6C_5=\dfrac{6!}{4!(6-4)!}\times4+6=66

The probability that committee at-least four women :

\dfrac{66}{210}=\dfrac{22}{70}

c) Number of ways that committee has more than 4 women :-

^6C_5\times^4C_0=6

The probability that committee has more than 4 women :-

\dfrac{6}{210}

Now, the  probability that committee has at most four women :-

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5 0
3 years ago
Is it parallel or is it perpendicular?
zavuch27 [327]
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and the slope is the coefficient of x 
so the slope = -5/4
in order to solve this problem we have to know the equation of the slop
slope=  \frac{y-y1}{x-x1}
where (x1,y1) is the point by which the new line passes through 

\frac{-5}{4}=\frac{y-(-4)}{x-1}
\frac{-5}{4}= \frac{y+4}{x-1}
-5(x-1)=4(y+4)
-5x +5=4y+16 (by subtracting 16 from both sides)
-5x -11= 4y (divide both sides by 4)
y= \frac{-5}{4}x - \frac{11}{4} 
4 0
3 years ago
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