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3 years ago

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Jonna is filling bags with candy. Each bag has 5 pieces of candy. Joanna wants 2/5 of the candy in each bag to be chocolate bars

. Solve using the table

2 answers:

lora16 [44]3 years ago

8 0

If she wants 2/5 she wants 2 pieces

Delvig [45]3 years ago

6 0

2 pieces of chocolate go into each bag

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Please Help!!!!!!! No stupid answers

Snowcat [4.5K]

A. The ratio of the person's shadow to their height is 6:2 or 3:1, so the height of the shadow is 360/3 or 120 m.
B. The model car's length divided by the ratio to the real car, which is 8.5/0.8 = 10.625 cm long.

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3 years ago

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

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

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

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

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

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

Part a:

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

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

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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,

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

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

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 $\frac{X}{4}$

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

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

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

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

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

7 0

3 years ago

No link or bot answer the question

sesenic [268]

Hello there! :)

$\sqrt{81}$

It's rational

it's an integer

Hope it helps!

~Just an emotional teen listening to her favorite song "Try Everything"

$SilentNature$

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2 years ago

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Which statement justifies that 3x² − 2x − 4 multiplied by 2x² + x − 3 obeys the closure property of multiplication?

Anna007 [38]

Step-by-step explanation:

(3x2 - 2x - 4) (2x2 + x - 3)

6x4 + 3x3 - 9x2 - 4x3 - 2x2 + 6x - 8x2 - 4x + 12

Solving like terms

6x4 - x3 - 19x2 + 2x + 12

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3 years ago

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36" = _____°<br><br> 0.01<br> 0.10<br> 0.50<br> 0.60

Tresset [83]

Answer:

.01

Step-by-step explanation:

There are 3600'' in 1°. 36"/3600"= .01°

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3 years ago