Giving brainiest to the correct answer

Mathematics

2 answers:

Nana76 [90]3 years ago

8 0

Answer:

Its A. 207 square cm

Step-by-step explanation:

Hope this helps!

____ [38]3 years ago

3 0

Answer:

207 sqaure cm

Step-by-step explanation:

also villagers are in the pic

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At 5:45 p.m., a jet is located 108 mi due east of a city. A second jet is located 214 mi due north of the city. To the nearest t

siniylev [52]

By connecting the distances traveled by the jet and the distance between them, we form a right triangle with hypotenuse equal to the unknown distance.
Using the Pythagorean,
h² = a² + b²
Substituting,
h² = (108 mi)² + (214 mi)²
h = 239.71 miles
Thus, the distance between them is approximately 239.71 miles.

7 0

4 years ago

Read 2 more answers

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

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).