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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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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below is the solution:

<span>distance = 475
</span>time = 9.5 hrs

<span>since distance and time varies directly
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therefore in 9.5 hrs train travels 475 miles => In 1 hr train travels 475 / 9.5 miles = 50 miles
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3 years ago
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In a random sample of fivefive microwave​ ovens, the mean repair cost was ​$75.0075.00 and the standard deviation was ​$13.0013.
Radda [10]
<h2>Answer with explanation:</h2>

As per given , we have

Sample size : n= 5

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Significance level for 90% confidence = \alpha=1-0.90=0.1

Using t-value table , t-critical value for 90% confidence:

t_{\alpha/2, df}=t_{0.05, 4}=2.132.

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Interpretation : The repair cost will be within $12.39 of the real population mean value \mu 90% of the time.

7 0
3 years ago
Ramen purchase a box of candy for himself on the way back to the dorm he ate 2/3 of the candy while he was putting away grocerie
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Answer:

12 chocolates

Step-by-step explanation:

Represent the original number of chocolates by c.  

Then (1/4)(2/3)c = 2

Solving this for c, we multiply both sides by 4:  (2/3)c = 8

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Originally the box contained 12 chocolates.

6 0
3 years ago
How do you solve it?
Elis [28]

At heart we're being asked for a line through two points,


(40^\circ \textrm{ C}, 355 \textrm { m/s}) \quad \textrm{and} \quad (49^\circ \textrm{ C}, 360 \textrm { m/s})


In general the line through (a,b) and (c,d) is


(y-b)(c-a)=(x-a)(d-b)


Check that you understand why both (a,b) and (c,d) are on this line.


Here our indepedent variable, instead of x, is T, temperature. Our dependent variable is v, velocity. Substituting,


(v - 355)(49 - 40) = (T - 40)(360 - 355)


9(v - 355) = 5(T - 40)


v-355 = \frac 5 9 T - \frac{200}{9}


v= \frac 5 9 T - \frac{200}{9} + 355


v= \frac 5 9 T + \frac{2995}{9}


That's our answer; let's check it.


When T=40, v = (5/9)40 + (2995/9) = 355 good


When T=49, v= (5/9)49 + (2995/9) = 360 good




7 0
2 years ago
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