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3 years ago

PLEASE HELP

2 answers:

8 0

You might want to plot these 2 points and draw a line segment connecting them. This line segment is the hypotenuse of a triangle. What are the lengths of the 2 shortest legs of this triangle? Square these lengths, and then sum up these two results. Find the square root of this sum. Round your answer to the nearest hundredth. Please show your work.

In summary, you could use either the distance formula or the Pythagorean Theorem to solve this problem. These two approaches are variations of one another.

sattari [20]3 years ago

7 0

Square root of 89 = 9.43398113 rounded to the nearest hundredth is 9.43 :)

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

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

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

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

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,

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

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 $\frac{X}{4}$

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

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:

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

The probability of no arrivals in a 15-second is 0.0821.

Part d:

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

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

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

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

7 0

3 years ago

Find the surface area of this right tringular Prism. Dimensions are in feet

larisa86 [58]

Ummm there’s no right triangular

5 0

3 years ago

Find all the angles listed. also list all of the lines that are secants, chords and tangents

Tanzania [10]

Answer:

Secants – 1. <CBG 2. <AGF 3. <ABD

Tangent – 1. <CDE

Chords – 1. <BD 2. <DF 3. <BG 4. <BF 5. <GD 6. <FG

Angles – 1. 45 2. 75 3. 35 4. 70 5. 75 6. 55 7. 50 8. 25 9. 35 10. 70 11. 50 12. 25 13. 70 14. 50. 15. 60 16. 85 17. 95 18. 85 19. 95

Step-by-step explanation:

4 0

3 years ago

Help see the screenshot below! !

Artemon [7]

Answer:

p = 8h

Step-by-step explanation:

for example, 16= 8 times 2 or 24 = 8 times 3

4 0

3 years ago

Read 2 more answers

Train A arrives at central station on the hour and every 12 minutes. Train B arrives o the hour and every 15 minutes. When do bo

lana [24]

Answer:

Every 30 hours

Step-by-step explanation:

Train A arrives at central station on the hour and every 12 minutes, you mean with that on every 72 minutes

Train B arrives at central station on the hour and every 15 minutes, you mean with that on every 75 minutes

In accordance with the given, the smallest number containing numbers 72 and 75 is 1800 minutes.

When we convert this number of minutes in hours we get:

1800 / 60 = 30 hours

God with you!!!

7 0

3 years ago