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VashaNatasha [74]

3 years ago

13

Evan has a shell collection. On Monday he found 6 new shells. On Tuesday , he gave 9 shells to his friends. After giving the she

lls away, Evan had 37 shells left. How many shells did Evan have to begin with?

1 answer:

Inessa05 [86]3 years ago

5 0

Answer:

52

Step-by-step explanation:

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Investors buy a studio apartment for 200,000. Of this amount, they have a down payment of 50,000. Their down payment is what pe

Mashcka [7]

Ummm I think it is 800,00

6 0

2 years ago

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

In a year group there 100 boys and 140 girls. Write as a ratio the number of boy to the number of girls. Give the answer in the

expeople1 [14]

no. of boys : no.of girls

100 : 140

to get the simplest ratio you gotta divide both with a common factor

Because both 100 and 140 share a common factor of 20

the simplest ratio will be

100/20 : 140/20

5:7

5:7 is the simplest ratio

5 0

2 years ago

Read 2 more answers

write the equaion of a line in slope-intercept form if the slope is 1/5 and the y-intercept is -9 explain please

faust18 [17]

Answer:

$y = \frac{1}{5} x - 9$

Step-by-step explanation:

The slope-intercept form is y = mx + b, where:

m = slope

b = y-intercept.

The slope (m) tells you the steepness of the line. It is the average rate of change which measures how the y-value changes for each one-unit change in the x-value. Hence, slope $= \frac{rise}{run}$ . So the given slope of 1/5 means that for every 1 unit change in the y-value, the x-value changes by 5 units (you go up 1 unit, and "run" 5 units to the right).

Next, the y-intercept is the point on the graph where it crosses the y-axis, and has the coordinates, (0, b). It is also the value of y when x = 0. Since you're given the y-intercept of -9, then that means that it is the y-coordinate of (0, <em>b</em>). So, it becomes (0, -9).

Now that we have our slope (<em>m </em>) = 1/5, and the y-intercept (<em>b </em>) = -9, we can write the equation of the line as:

$y = \frac{1}{5} x - 9$

(I'm also including a screenshot of the line where it shows the y-intercept of (0, -9) on the graph).

Please mark my answers as the Brainliest if you find my explanations helpful :)

7 0

2 years ago

Solve thank you I will give brainlist to first

Gre4nikov [31]

Answer:

Step-by-step explanation:

no because each y value doesn’t increase uniformly

4 0

3 years ago

Read 2 more answers