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White raven [17]
2 years ago
6

Suppose the sample space for a probability experiment has 8 elements. If elements from the sample are selected without replaceme

nt, how many different ways can you select all of them? Remember that "without replacement" means that the items are not returned to the sample space after they are chosen. Write your answer in factorial notation.
Mathematics
1 answer:
Gwar [14]2 years ago
5 0

Answer: Number of ways =\dfrac{8!}{8!0!}\ or \ 1

Step-by-step explanation:

When there is no replacement , we use combination to find the number of ways to choose things.

Number of ways to choose r things out of  ( with out replacement) = ^nC_r=\dfrac{n!}{r!(n-r)!}

The number of ways to choose  8 things = ^8C_8=\dfrac{8!}{8!(8-8)!}=\dfrac{1}{0!}=1  [\because 0!=1]

Hence,  Number of ways =\dfrac{8!}{8!0!}\ or \ 1

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Chapter : Circles 1 question
erica [24]
We can solve this question in 2 ways: either using degrees or converting the degrees into radians.

Since, the question says degrees itself and there is no specification of using radians only, so I have solved it using degrees itself.

Part (a):

Perimeter of sector ORS = 2*Radius + Arc RS = 2*21 + 60*  \frac{ \pi }{180} *21 =  \frac{ \pi }{3} *21 = 7 \pi

Part (b):

Area of sector ORS = \frac{60}{360} * \pi * 21^{2} =  \frac{1}{6} * \pi *21*21=  \frac{147}{2} \pi

Area of sector POQ = \frac{36}{360} *  \pi * 14^{2} =  \frac{1}{10} *  \pi * 196 =  \frac{196}{10}  \pi

Thus, area of shaded region
= Area of sector ORS - Area of sector POQ

= \frac{147}{2} \pi - \frac{196}{10} \pi =  \frac{735 - 196}{10}  \pi =  \frac{539}{10}  \pi = 53.9 \pi

5 0
3 years ago
A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble. HINT (See Example 7.] H
Murljashka [212]

Answer:  35

Step-by-step explanation:

Given : A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble.

Total = 4+2+1+3+1=11

To find sets of four marbles include none of the red ones, we need to exclude red marbles when we count the total number of marbles.

Then, the total marbles(exclude red) =11-4=7

Now, the combination of 7 marbles taking 4 at a time is given by :-

^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!3!}=35

Hence, the number of sets of four marbles include none of the red ones = 35

4 0
3 years ago
The number of students in a chess club decreased from 27 to 11. What is
xz_007 [3.2K]

Answer:

59.26.

Step-by-step explanation:

I think!

7 0
3 years ago
Please help ill give brainliest pls dont guess
LiRa [457]
When y=2 and y=5

1. 2y-1 and (3y-5+y or 4y-5)
when y=2 ; 2(2)-1 = 3 and 4(2)-5=3
when y=5 ; 2(5)-1 = 9 and 4(5)-5=15

----nonequivalent-----

2.5y+4 and (7y+4-2y or 5y+4)
so you don't have to place any value in because 5y+4 and 7y+4-2y are equal,
whatever you place any value in, it will be all the same then

-----equivalent------

and no need to find more
7 0
3 years ago
Yallll please help me its my next period and its missing.​
jeka57 [31]

Answer:

<em>#6: -1/5     #7: 2/3</em>

Step-by-step explanation:

#6:

(5,7) and (25,3)

3-7/ 25-5 = -1/5

#7:

( 6,4) and (12,8)

8-4/ 12-6 = 2/3

i hope this helped! :D

5 0
3 years ago
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