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Volgvan
2 years ago
9

Find the values of x (there are two) when the ratio of the area of the shaded portion of the figure to the total area of the fig

ure is 15.75/48.

Mathematics
1 answer:
Arada [10]2 years ago
6 0

The values of x are x = -0.77 and x = 3.27

<h3>How to determine the value of x?</h3>

To calculate the value of x. we start by using the following equivalent ratio

2x : (3.5 + x + 1.5) = x + 0.5 : x + 1.5

Evaluate the like terms

2x : (5 + x) = x + 0.5 : x + 1.5

Express the ratio as a fraction

2x/5 + x = x + 0.5/x + 1.5

Cross multiply

2x(x + 1.5) = (5 + x)(x + 0.5)

Open the brackets

2x^2 + 3x = 5x + 2.5 + x^2 + 0.5x

Evaluate the like terms

x^2 - 2.5x - 2.5 = 0

Using a graphing calculator, we have

x = -0.77 and x = 3.27

Hence, the values of x are x = -0.77 and x = 3.27

Read more about triangles at:

brainly.com/question/17335144

#SPJ1

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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
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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

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