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3241004551 [841]
3 years ago
13

Angelo reads 1&1/2 pages in 3 minutes. Find his reading speed in pages per hour

Mathematics
2 answers:
dedylja [7]3 years ago
8 0

Answer:

30 pages per hour

Step-by-step explanation:

Angelo reads 1&1/2 pages in 3 minutes.

1 \frac{1}{2} =\frac{3}{2}

3/2 pages read in 3 minutes, we need to find the pages read in 1 hour

we know that 1 hour = 60 minutes

In 3 minutes pages read = 3/2

To find out reading speed, we divide number of pages by minutes

Reading speed per minute= \frac{\frac{3}{2} }{3} =\frac{1}{2}

So his reading speed in pages per hour is \frac{1}{2} \cdot 60= 30

30 pages per hour

Dimas [21]3 years ago
3 0

Answer:

30 pages a hour

Step-by-step explanation:

1.5 every 3 minutes

there is 60 minuts in an hou you already have 3 so thimes it by 20 to get 60 then times 60 by 1.5


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In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p
skelet666 [1.2K]

Answer:

\mathbf{P(X=5) =0.0888}    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

\mathtt{P(X=x) =(^{n}_{x} )   \  \pi^x \  (1-\pi)^{n-x}}

\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} )   \  \pi^x \  (1-\pi)^{n-x}}

where;

n = 8 and π = 0.36

For x = 5

The probability \mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} )   \  0.36^5 \  (1-0.36)^{8-5}}

\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} )   \  0.36^5 \  (0.64)^{3}}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =({8 \times 7 } )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =0.0887645}

\mathbf{P(X=5) =0.0888}     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})

{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times  (0.36)^0  \times  (1-0.36)^8  \ )  +  \dfrac{8!}{1!(7!)} \times  (0.36)^1  \times  (1-0.36)^7  \ +\dfrac{8!}{2!(6!)} \times  (0.36)^2  \times  (1-0.36)^6  \ +  \dfrac{8!}{3!(5!)} \times  (0.36)^3  \times  (1-0.36)^5 +  \dfrac{8!}{4!(4!)} \times  (0.36)^4  \times  (1-0.36)^4  \  +  \dfrac{8!}{5!(3!)} \times  (0.36)^5  \times  (1-0.36)^3  \ )

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

4 0
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Hi everybody! I need help, and this question is worth TEN points!
harkovskaia [24]

Answer:


Step-by-step explanation:

we know,

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Here, W is the internal point.

The two angles are  UVW and WVX.

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UVX is equal to the sum of  UVW and WVX.

Hence, the reason is-

Angle Addition Postulate

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