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3 years ago

Answers in Scientific Notation.

Mathematics

1 answer:

dmitriy555 [2]3 years ago

7 0

one way is to factor the expressions

(a)

$10^{5}$ (1.2 + (5.35 × $10^{1}$ ))

= $10^{5}$ (1.2 + 53.5)

= 54.7 × $10^{5}$ = 5.47 × $10^{1}$ × $10^{5}$ = 5.47 × $10^{6}$

(b)

$10^{-3}$ ((6.91 × $10^{1}$ ) + 2.4)

= $10^{-3}$ (69.1 + 2.4)

= 71.5 × $10^{-3}$ = 7.15 × $10^{1}$ × $10^{-3}$ = 7.15 × $10^{-2}$

(c)

$10^{5}$ ((9.7 × $10^{1}$ ) + 8.3)

= $10^{5}$ (97 + 8.3)

= 105.3 × $10^{5}$ = 1.053 × $10^{2}$ × $10^{5}$ = 1.053 × $10^{7}$

(d)

$10^{1}$ ((3.67 × $10^{1}$ ) - 1.6)

= $10^{1}$ (36.7 - 1.6)

= 35.1 × $10^{1}$ = 3.51 × $10^{1}$ × $10^{1}$ = 3.51 × $10^{2}$

(e)

$10^{-6}$ ((8.41 × $10^{1}$ ) - 7.9)

= $10^{-6}$ (84.1 - 7.9 )

= 76.2 × $10^{-6}$ = 7.62 × $10^{1}$ × $10^{-6}$ = 7.62 × $10^{-5}$

(f)

$10^{4}$ ((1.33 × $10^{1}$ ) - 4.9 )

= $10^{4}$ (13.3 - 4.9) = 8.4 × $10^{4}$

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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

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