There are 91 such ways in whih the volunteers can be assigned if two of them cannot be assigned from 14 volunteers.
Given that a school dance committee has 14 volunteers and each dance requires 3 volunteers at the door, 5 volunteers on the floor and 6 on floaters.
We are required to find the number of ways in which the volunteers can be assigned.
Combinations means finding the ways in which the things can be choosed to make a new thing or to do something else.
n
=n!/r!(n-r)!
Number of ways in which the volunteers can be assigned is equal to the following:
Since 2 have not been assigned so left over volunteers are 14-2=12 volunteers.
Number of ways =14
=14!/12!(14-12)!
=14!/12!*2!
=14*13/2*1
=91 ways
Hence there are 91 such ways in whih the volunteers can be assigned if two of them cannot be assigned.
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Answer:
4=????????????
Step-by-step explanation:
Answer:
Prob ( 41 <= X <= 77) =
Prob ( (41-59)/9 <= Z <= (77-59)/9 ) = <--- changes to z scores
Prob ( -18/9 <= Z <= 18/9 ) =
Prob ( -2 <= Z <= 2) =
Prob ( z<=2) - Prob(Z<=-2) = <--- by property of the normal bell curve
= 0.9772 - 0.0228 <--- per the normal distribution table
= 0.9544
Step-by-step explanation:
make sense ?
The price of a senior citizen ticket is $8 and the price of a student ticket is$ 10
Step-by-step explanation:
Let the cost of one seniors citizen ticket be x and student be y
First day:
10 seniors citizen tickets and 11 student for a total of $190.
10x + 11y = 190-----------------------------(1)
Second day:
$160 on the second day by selling 5 senior citizen tickets and 12 student tickets.
5x + 12y = 160---------------------------------(2)
Multiply eq(2) by 2
10x + 24y = 320----------------------------(3)
Subtract eq(1) from (3)
10x + 24y = 320
10x + 11y = 190
-----------------------------
0x + 13 y = 130
---------------------------
13y = 130

y = 10-------------------------------(4)
Substituting (4) in (2)
5x + 12(10) = 160
5x + 120 = 160
5x = 160-120
5x = 40

x = 8
Yes because she does have 21.99 left good luck man