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3 years ago

The following table shows the time in various cities when it is 4:00 a.m. in New York.

2 answers:

8 0

It's Dublin - San Juan

Departure Friday at 1:30 Pm (Dublin Time) ==> Time San Juan 9:30 (-4 Hours)

Travel time 16 Hrs ( Dub Sat at 6:30PM) & arriving San JUan on Sat at 9:30 +16;30 26 that means 2 AM (26-24)

Diano4ka-milaya [45]3 years ago

3 0

Answer:

The correct option is C

Step-by-step explanation:

Let the flight go between X and Y

Time at which flight leaves X is 1:30 p.m (local) and time at which flight arrives in Y is 2:00 a.m (local)

The time in city X when the flight arrives in Y = 1:30 p.m (friday) + 16 hours 30 minutes

= 6 a.m (saturday). But the local time of Y is given 2 a.m

So, the flight will go between those two cities where time in of first is 4 hours ahead of time in second

And according to options, C. is correct because when the time is 6 a.m. in Dublin the time in San Juan is 2 a.m which is our required cities X and Y

So, the flight will go between X and Y

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2+2 write near doubles for double facts. Write the sum

Rashid [163]

Answer: the sum of 2+2? Ok. The answer is 4. If it's not the answer then I am truly sorry.

Step-by-step explanation:

4 0

3 years ago

53% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

faltersainse [42]

Answer:

$P(X = 5) = 0.2423$

$P( X \ge 6 )= 0.4516$

$P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.53

The sample size is n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as

$P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=> $P(X = 5) = 252 * 0.04182 * 0.023$

=> $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

$P( X \ge 6 )= P( X = 6) + P( X = 7) + P( X = 8) + P( X = 9) + P( X = 10)$

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

$P( X < 4 )= P( X = 3) + P( X = 2) + P( X = 1) + P( X = 0)$

=> $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$