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3 years ago

Help please, i need this done asap thanks in advance

Mathematics

1 answer:

alina1380 [7]3 years ago

7 0

Answer:

<AOC

<AOB

<BOC

Step-by-step explanation:

The central angles are formed at point O, which is the center of the circle.

All the central angles are marked in the attachment.

The central angles are

: $m\angle AOC$ . This is a right angle and it is 180 degrees.

: $m\angle AOB$

: $m\angle BOC$

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United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

a) The probability that exactly 4 flights are on time is equal to 0.0313

b) The probability that at most 3 flights are on time is equal to 0.0293

c) The probability that at least 8 flights are on time is equal to 0.00586

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures.

a) The probability that exactly 4 flights are on time is =

b) The probability that at most 3 flights are on time is =

c)The probability that at least 8 flights are on time is =

<h2>Solution to the problem</h2>

a) Probability that exactly 4 flights are on time

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

$_nC_x=\frac{x!}{n!(n-x)!}$

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

b) Probability that at most 3 flights are on time

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

c) Probability that at least 8 flights are on time 

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

3 years ago

Please answer correctly !!!! Will mark brainliest !!!!!!!

enyata [817]

Answer:

Step-by-step explanation:

5*5=25

25-3=22

5 0

3 years ago

BIG POINTS FOR ANYONE WHO SOLVES DIS PLZ

vovikov84 [41]

The answer is

${36x}^{2} + 12x + 1$

If we put 36x^2, then the answer we would get is

${(6x + 1)}^{2}$

Hope this helps :)

6 0

2 years ago

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Please help 30 points

Ulleksa [173]

Replace x and y with the given values:

2(1/8 + 4) - (3/16*8)

Use distributive property:

1/4 + 8 - (3/16 * 8)

Simplify to get final answer:

8 1/4 - 1 1/2 = 6 3/4

4 0

3 years ago

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I need help please do this for me 49 times 3 divided by 7 square it times 1/4 Add 3/4

mylen [45]

Answer:

3/2 or 1.5 or 1 1/2

Step-by-step explanation:

8 0

3 years ago