Ask question

Ask question

All categories

adelina 88 [10]

3 years ago

7

an aircraft flew 5 hours with the wind. the return trip took 6 hours against the wind. if the speed of the plane in still air is

220 miles per hour more than the speed of the wind, find the wind speed and the speed of the plane in still air

1 answer:

olga2289 [7]3 years ago

7 0

w = 37.846 mph (wind rate)
plane speed in still air: 492 mph

You might be interested in

Please help due tonight !!!!

gregori [183]

Answer:

okay where is the question

Step-by-step explanation:

follow me on ig dipeshkumar0853822

and on brainly also

4 0

3 years ago

What is the value of f(3) when f(x) = 4x + 1

valentina_108 [34]

Here is the equation:

$f(x)=4x+1$

You need to find f(3). To do that, substitute all the x's with 3:

$f(x)=4x+1 \rightarrow f(3)=4\times3 + 1$

The first thing you need to do is multiply 4 by 3. Multiply:

$4 \times 3 = 12$

Here is your new equation:

$f(3)=12+1$

To get your answer, you need to add 12 and 1. Add:

$12+1 = 13$

That means:

$\bf f(3)=13$

<em>If you have any questions, feel free to ask in the comments! :)</em>

3 0

3 years ago

Read 2 more answers

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

Write an equivalent expression.

djyliett [7]

8.5 the answer is and replace the areas

4 0

3 years ago

An aircraft factory manufactures airplane engines. The unit cost (the cost in dollars to make each airplane engine) depends on t

Juliette [100K]

Answer:

Minimum unit cost = 5,858

Step-by-step explanation:

Given the function : C(x)=x^2−520x+73458

To find the minimum unit cost :

Take the derivative of C(x) with respect to x

dC/dx = 2x - 520

Set = 0

2x - 520

2x = 520

x = 260

To minimize unit cost, 260 engines must be produced

Hence, minimum unit cost will be :

C(x)=x^2−520x+73458

Put x = 260

C(260) = 260^2−520(260) + 73458

= 5,858

7 0

3 years ago