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algol13
3 years ago
10

Ten helicopters are assigned to search for a lost airplane; each of the helicopters can be used in one out of two possible regio

ns where the airplane might be with the probabilities 0.8 and 0.2. How should one distribute the helicopters so that the probability of finding the airplane is the largest if each helicopter can find the lost plane within its region of search with the probability 0.2, and each helicopter searches independently? Determine the probability of finding the plane under optimal search conditions.
Mathematics
1 answer:
OLga [1]3 years ago
3 0

Answer:

flew there at the game

Step-by-step explanationit can change

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What strategies do you use when solving for an unknown number​
frez [133]

Answer:

adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.

Step-by-step explanation:

please mark brainleist

5 0
2 years ago
1/2(6+3p)-3/4(8-10p)<br> Expand and simplify
andreev551 [17]
When you take this problem and simplify it you get 9p-3
5 0
3 years ago
Read 2 more answers
In a set of 25 aluminum castings, four castings are defective (D), and the remaining twenty-one are good (G). A quality control
lianna [129]

Answer:

The sample space for selecting the group to test contains <u>2,300</u> elementary events.

Step-by-step explanation:

There are a total of <em>N</em> = 25 aluminum castings.

Of these 25 aluminum castings, <em>n</em>₁ = 4 castings are defective (D) and <em>n</em>₂ = 21 are good (G).

It is provided that a quality control inspector randomly selects three of the twenty-five castings without replacement to test.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

{n\choose k}=\frac{n!}{k!(n-k)!}

Compute the number of samples that are possible as follows:

{25\choose 3}=\frac{25!}{3!\times (25-3)!}

      =\frac{25\times 24\times 23\times 22!}{3!\times 22!}\\\\=\frac{25\times 24\times 23}{3\times 2\times 1}\\\\=2300

The sample space for selecting the group to test contains <u>2,300</u> elementary events.

6 0
3 years ago
A number of friends decided to go on a picnic and planned to spend rs. 96 on eatables. four of them, however, did not turn up. a
mr Goodwill [35]

The number of those who attended the picnic was 8.

Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax^{2} + bx + c = 0 where a, b, c, ∈ R and a ≠ 0.

Such questions are best examples of quadratic equations where  the unknown is simply assumed to be a variable which when substituted according to the conditions yield a quadratic equation which can easily be solved.

Let x represent the whole population.

Total Spending = Rs. 96

Thus, each individual's contribution is equal to 96/x.

Four people did not show up causing

others to contribute Rs. 4 extra.

So the given condition can be written as

96/(x-4) – 96/x = 4

=>96x – 96(x-4) = 4x(x-4)

=> 96x -96x + 384 = 4x2 – 16x

=> 4x2 – 16x – 384 = 0

=> x2 – 4x – 96 = 0

=> (x – 12)(x + 8) = 0

=> x = 12 or x = -8 (neglect)

So x = 12

x-4 = 8  ( because  four of them did not turn up)

∴ The number of those who attended the picnic was 8.

Learn more about quadratic equations here :

brainly.com/question/15241362

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4 0
2 years ago
You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually. What would the value of the car be in 20
Fed [463]

Answer: -$6,500

Step-by-step explanation:

Here we could , use the arithmetic progression where

T(2020 - 2010) = a + ( n - 1 )d

T10 = a + ( 10 - 1 )d --------------- 1

a = $25,000, n = 10 and d = 14% of $25,000 = $3,500 the common difference.

Note since it decreases the common difference d = -$3,500.

Now substitute for the values in the equation above.

T10 = 25,000 + 9 x -3,500

= $25,000 - $31,500

= -$6,500 (deficit )

4 0
3 years ago
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