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iragen [17]
3 years ago
9

Solve using any strategy. A school band has a brass section of trumpet, trombone, and tuba players. There are twice as many trom

bones as tubas, and half as many trombones as trumpets. If there are two tubas in the band, what is the total number of players in the brass section
A. 8 players


B. 10 players


C. 12 players


D. 14 players
Mathematics
2 answers:
Roman55 [17]3 years ago
7 0
2 Tubas, 4 Trumpets, 8 Trombones, all together you have 14 Players. The answer is D
never [62]3 years ago
6 0
If there are 2 tubas and there are twice as many trombones as there are tubas then there are 4 trombones. And if there are half as many trombones as there are trumpets, then there are 8 trumpets. Therefore, the answer is (D).
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40. Find UW.<br> T<br> 6<br> 5<br> U +-<br> The measure of UW is<br> type your answer...
Ahat [919]

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9

Step-by-step explanation:

UT^2 = UV * UW

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7 0
3 years ago
(3 points) Blades of grass Suppose that the heights of blades of grass are Normally distributed and independent, with each heigh
NemiM [27]

The final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

What is the normal distribution?

A continuous probability distribution for a real-valued random variable in statistics is known as a normal distribution or Gaussian distribution.

If x follows a normal distribution with mean μ and standard deviation σ then the distribution of

\sum_{i =1}^{n}x_{i}  follows an approximately normal distribution with a mean n\mu and standard deviation \sqrt{n }\sigma

let x be the height of blades of grass

x follows normal distribution with mean = μ = 4 and standard deviation = σ = 0.75.

Y = x1 + x2 +...........+x10

Y = \sum_{i =1}^{10}x_{i}

Distribution of Y is normal with,

Mean = \mu _{y}=10*4 = 40 and standard deviation = \sigma _{y}=\sqrt{10}*0.75 = 2.3717

a)

P( Y < 42.5 )

Using normal distribution formmula,

f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}

=NORMDIST( x, mean, SD , 1 )      

=NORMDIST(42.5, 40, 2.3717, 1 )

=0.8541

P( Y < 42.5 )  = 0.8541

b)

P( 39.5 < Y < 40.5 ) = P( Y < 40.5 ) - P( Y < 39.5 )

Using normal distribution formmula,

f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}

P( Y < 40.5 )  =NORMDIST(40.5, 40, 2.3717, 1 ) = 0.5835

P( Y < 39.5 ) = NORMDIST(39.5, 40, 2.3717, 1 ) = 0.4165

P( 39.5 < Y < 40.5 ) = 0.5835 - 0.4165  = 0.1670

P( 39.5 < Y < 40.5 ) = 0.1670

Hence, the final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

To learn more about the normal distribution visit,

brainly.com/question/4079902

#SPJ4

5 0
1 year ago
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