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Tema [17]
3 years ago
11

A plane traveled 1248 miles to Ft. Worth and back. The trip there was with the wind. It took 12

Mathematics
1 answer:
Artemon [7]3 years ago
6 0

speed of the  plane in still air is 76 m/s and speed of the wind is 28 m/s .

<u>Step-by-step explanation:</u>

Here we have , A plane traveled 1248 miles to Ft. Worth and back. The trip there was with the wind. It took 12  hours. The trip back was into the wind. The trip back took 24 hours. We need to find What is the speed of the  plane in still air? What is the speed of the wind? Let's find out :

<u>For Going trip :</u>

Let speed of plane be u , and wind be v so : Speed = ( Distance ) / ( time )

⇒ u+v = \frac{1248}{12}

⇒ u+v = 104   ...........(1)

<u>For Returning Trip :</u>

⇒ u-v = \frac{1248}{24}

⇒ u-v = 52   ...........(2)

Adding both equations we get :

⇒ (u+v)+(u-v)=104+52

⇒ 2u=156

⇒ u= 76m/s

Putting this value in equation (1) we get :

⇒ 76+v = 104

⇒ v=28m/s

Therefore , speed of the  plane in still air is 76 m/s and speed of the wind is 28 m/s .

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Answer:

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Step-by-step explanation:

32-x-x-x-x-x-x-x

32/4 = 8

so product 1/4 is the rule

32/4^5= 2^5/2^10

2^-5=1/32

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2 years ago
Solve this system algebraically.
seraphim [82]
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3 years ago
A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has
bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] =  1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

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3 years ago
Jacie can swim 3/4 of a mile in 3/8 of an hour.how far can she swim in one hour?
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Answer:

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Step-by-step explanation:

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If f(x)=-3 then x=___
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Answer:

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Step-by-step explanation:

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