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OlgaM077 [116]
3 years ago
9

Aman and Prakash coloured an entire Harry Potter Colouring Book together. The book had 75 pages. Aman coloured 60% of the pages.

How many pages did Prakash colour? A) 30 b) 60 c) 75 d) 45
Mathematics
2 answers:
tia_tia [17]3 years ago
8 0

Answer:

A) 30 pages

Step-by-step explanation:

No of pages = 75

Aman coloured 60 % of the pages= 75 *60/100= 45

Prakash coloured 75 - 45 = 30 pages

Scilla [17]3 years ago
6 0

Answer: Prakash coloured 30 pages of the book.

Step-by-step explanation:

The total number of pages in the entire Harry Potter Colouring Book is 75. Aman and Prakash coloured an entire Harry Potter Colouring Book together. Aman coloured 60% of the pages. This means that the percentage of the entire Harry Potter Colouring Book that Prakash coloured would be 100 - 60 = 40%

The number of pages that Prakash coloured would be

40/100 × 75 = 0.4 × 75 = 30 pages

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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
If the width of the rectangle increases by 4 and the length decreases by 3, by how much will the area change? Express the differ
OlgaM077 [116]

Answer:

this might be the answer..

Step-by-step explanation:

m²+6m+9

3 0
3 years ago
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