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Tju [1.3M]
4 years ago
10

Hi, can someone please solve this question a put it on a number line to help me out? 0

Mathematics
2 answers:
DENIUS [597]4 years ago
8 0

Answer:whats the question

Step-by-step explanation:

lara31 [8.8K]4 years ago
3 0
What’s the question?
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20% of 59.99 is 11.998 so 11.99. 59.99 - 11.99 = 48. 48x4 is 192
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Nikitich [7]

The function is going downward

Answer is D.

As x approaches negative infinity, the function approaches negative infinity.

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Match the following reasons with the statements in the given proof. Prove that if two angles are complementary to the same angle
yulyashka [42]

1. --> d

2. --> b

3. --> a

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1 is d because that is the only piece of information that sounds like a given.

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3 is a because since both of the problems equal 90 degrees, you can just take away the 90 and put an equal sign in between the two problems because they equal the same thing.

4 is c because you are subtracting m∠2 from the problems and ending up with just m∠1 = m∠3.

Hope this helps! :)

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3 years ago
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(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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