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Allisa [31]
3 years ago
12

The school band is scheduled to march in the annual parade, but the band's starting point is changed the day before the parade.

The band director calls three band members. Each band member calls three other band members. Then these band members each call three members. How many band members, including the band director, are notified of the new starting point?
Question 8 options:

20 members

12 members

13 members

40 members

Mathematics
1 answer:
Stels [109]3 years ago
5 0

Answer:

There are 40 band members including the band director that were notified of the new starting point

Step-by-step explanation:

The diagram below shows the band director at the top, then the three band members he called, then the next band members, and so forth.

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Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $3,159 was collected on the sale of 1,339 ticket
ivolga24 [154]

Answer:

  • 455 adult tickets
  • 884 student tickets

Step-by-step explanation:

Let "a" represent the number of adult tickets sold. Then 1339-a is the number of student tickets sold. The revenue collected was ...

  5a + 1(1339-a) = 3159

  4a = 1820 . . . . . . subtract 1339, collect terms

  a = 455 . . . . . . . . divide by 4; adult tickets sold

  (1339-a) = 884 . . . . student tickets sold

455 adult tickets and 884 student tickets were sold.

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3 years ago
Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answ
gulaghasi [49]

Answer:

Part a ) The degrees of freedom for the given two sample non-pooled t test is 24

Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf =  {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf  = 30184 / 1241

Δf  = 24.3223 ≈ 24 (down to the nearest whole number)

b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

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Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41  

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

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