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rodikova [14]
3 years ago
15

If it takes 6 hours for a plane to travel 720km with a tail wind and 8 hours to make the return trip with a head wind. Find the

air speed of the plane and speed of the wind.
Mathematics
1 answer:
qaws [65]3 years ago
8 0

Answer:

The air speed of the plane is 90 km/hr. and the wind has a speed of 15 km/hr.

Step-by-step explanation:

The plane travels 720 km in 6 hours in the direction of the wind and it takes 8 hours to make the round trip.

If we assume that the speed of the plane is u km/hr. and that of wind is v km/hr.

Therefore, we can write u + v = \frac{720}{6} =120 ........ (1)

And u-v = \frac{720}{8}=90 ..... (2)

If we add those two equations, then 2u = 210, ⇒ u = 105 km/hr.

Again from equation (2), we get v = u - 90 = 105 - 90 = 15 km/hr.

Therefore, the airspeed of the plane is 90 km/hr. and the wind has a speed of 15 km/hr. (Answer)

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An archaeologist descends 20 feet into a canyon, then climbs up 12 feet. Which value represents her final position?
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-20 + 12 = -8 feet descent.
4 0
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Look at the figure. A 10 foot tall flagpole cast a shadow that is 24 feet long what is the approximate length of the shadow cast
masya89 [10]
Approx 60 feet see pic for work

7 0
3 years ago
6. Christopher can make enough iced coffee for one
Elis [28]

Answer:

B) 16

Step-by-step explanation:

<u>Espresso required for one person:</u>

  • 2 teaspoons

<u>Espresso required for 24 people:</u>

  • 2*24 = 48 teaspoons

3 teaspoons = 1 tablespoon

48 teaspoons = 48/3 = 16 tablespoons

Christopher needs 16 tablespoons of espresso.

<u>Correct answer choice:</u> B) 16

4 0
3 years ago
If m&lt;FHI=142, what are m&lt;FHG and m&lt;GHI​
lisabon 2012 [21]
<h3>Answers:</h3>

angle FHG = 42 degrees

angle GHI = 100 degrees

=========================================================

Explanation:

FHI is the largest angle. It is split into two pieces FHG and GHI which have measures of (3x+6) and (9x-8) degrees respectively.

Put another way, those two smaller angles (FHG and GHI) combine to form the larger angle (FHI)

This is the angle addition postulate.

------------------

(angle FHG) + (angle GHI) = angle FHI

(3x+6) + (9x-8) = 142

3x+6+9x-8 = 142

(3x+9x) + (6-8) = 142

12x - 2 = 142

12x-2+2 = 142+2 ... adding 2 to both sides

12x = 144

12x/12 = 144/12 .... dividing both sides by 12

x = 12

------------------

Use that x value to find the angles we're after

angle FHG = 3x+6 = 3*12+6 = 36+6 = 42

angle GHI = 9x-8 = 9*12-8 = 108-8 = 100

Note how

(angle FHG) + (angle GHI) = 42 + 100 = 142

which is the measure of angle FHI. This confirms the answers.

8 0
2 years ago
Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g
Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

    $E[g(x)] = \sum g(x)p(x)$

So,  $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

       $g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also,  $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

   $g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0
2 years ago
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