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3 years ago

Please solve. 6(3 + 8) - 4/5

Mathematics

2 answers:

Rudik [331]3 years ago

7 0

Answer:

$\Large \boxed{\frac{326}{5}}$

Step-by-step explanation:

$\displaystyle 6(3 + 8) - \frac{4}{5}$

Solving brackets.

$\displaystyle 6(11) - \frac{4}{5}$

$\displaystyle 66 - \frac{4}{5}$

Subtract.

$\displaystyle \frac{66}{1} - \frac{4}{5}$

$\displaystyle \frac{66 \times 5}{1 \times 5} - \frac{4}{5}$

$\displaystyle \frac{330}{5} - \frac{4}{5}$

$\displaystyle \frac{330-4}{5}$

$\displaystyle \frac{326}{5}$

Gre4nikov [31]3 years ago

3 0

G'Day mate! Here's the answer:

53 1/3

1. 6(3 + 6) - 4/5

2. 6 x 3 is 18, and 6 x 8 is 48

3. 18 plus 48 is 66

4. 66 - 4/5 = 266/5

5. 266/5 as mixed number is 53 1/5

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Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

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$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

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The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

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The probability of at least one arrival in a 15-second period is 0.9179.

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Answer:

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