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3 years ago

................. plz let only deltaalex answer:) don’t trust anyone on here but them!

Mathematics

1 answer:

nevsk [136]3 years ago

3 0

Answer:

Step-by-step explanation:

oh law of sines soo sin(A) / a = Sin(B) / b

2a)

here DE = side f then let's find that side with law of sines and plug side f in as 'a' in law of sines , so

Sin(28) / a = Sin(54) / 31 ( where a = f = DE )

can you do the algebra here Emily? or do you want me to show that?

Sin(28) *31 / Sin(54) = a

17.989 =a

18.0 =a (rounded to nearest 10th)

DE = 18.0 m

2b)

find RT where RT = side a

since they give use 2 of the other sides of the triangle we can quickly figure out the S angle. 180 = 71+62+S so S = 47° ( I'm going kinda fast b/c I don't know how much time you have :? )

then Sin(A) /a = Sin(B) / b

Sin(47) / a = Sin(62) / 29

Sin(47) *29 / Sin(62) = a

24.020 = a

RT = 24.0 meters ( rounded to nearest 10th )

2c)

they want us to find ∠X with law of sines soo

I'll just use the letter they give us for variables

Sin(X) / x = Sin(Y ) / y

Sin(X) = x* Sin(y) / y

X = arcSin [ 14 * Sin(108) / 34 ]

X = arcSin [ 0.3916115]

X = 23.0548°

X= 23° ( rounded to nearest degree )

2d)

they ask us to find ∠E

I'll , again, use the letter they give us for variables

Sin(E) / e = Sin( F ) / f

Sin(E) / 18 = Sin(78) / 30

E = arcSin[ 18 * Sin(78) / 30 ]

E = arcSin [ 0.5868856 ]

E = 35.936 °

E = 36 ° ( rounded to nearest degree )

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The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So

pshichka [43]

Answer:

There is a 32.22% probability that at least 3 flights arrive late.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrives on time, or it arrives late. This means that we can solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

There are 10 flights, so $n = 10$ .

A success in this case is a flight being late. 80% of its flights arriving on time, so 100%-80% = 20% arrive late. This means that $\pi = 0.2$ .

(a) Find the probability that at least 3 flights arrive late.

Either less than 3 flights arrive late, or at least 3 arrive late. The sum of these probabilities is decimal 1. This means that:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

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

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6778 = 0.3222$

There is a 32.22% probability that at least 3 flights arrive late.

3 0

3 years ago

It took a plane 2 hours to go from Dallas to NY with the head wind and 1.5 hours to get back with the tail wind. If the speed of

vladimir2022 [97]

Answer:

360 miles

Step-by-step explanation:

let speed of plane be x

speed of wind be c

Also note D = RT, where

D is distance and R is rate and T is time

For First Leg:

we can write:

(x - c) * 2 = D

2x - 2c = D

For Second Leg:

we can write:

(x+c)*1.5 = D

1.5x + 1.5c = D

We can equate and write:

2x - 2c = 1.5x + 1.5c

0.5x = 3.5c

We know Speed of Wind, c, is 30, so we have:

0.5x = 3.5(30)

0.5x = 105

x = 105/0.5 = 210 [speed of airplane]

We know distance is

D = 2x - 2c

D = 2(210) - 2(30)

D = 420 - 60

D = 360 miles

The distance is 360 miles

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Answer:

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