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2 years ago

6

Amber needs to select six players for her volleyball team. there are 10 players to choose from. how many possible combinations o

f teams could amber choose?

1 answer:

Dima020 [189]2 years ago

8 0

Amber can make a possible number of 12 different teams

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A

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2 years ago

What is the value of 24/25 divided by 4/5

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1.2

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3 years ago

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The exponential distribution applies to lifetimes of a certain component. Its failure rate is unknown. Find the probability that

wlad13 [49]

Answer:

a) 0.0821

b) 0.0111

c) 0.0041

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \lambda e^{-\lambda x}$

In which $\lambda$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\lambda x}$

Either it lasts more 5 years or less, or it survives more than 5 years. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

In all 3 cases, we want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = 1 - e^{-5\lambda}$

(a) lambda=.5

$P(X \leq 5) = 1 - e^{-5*0.5} = 0.9179$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9179 = 0.0821$

(b) lambda=0.9

$P(X \leq 5) = 1 - e^{-5*0.9} = 0.9889$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9889 = 0.0111$

(c) lambda=1.1

$P(X \leq 5) = 1 - e^{-5*1.1} = 0.9959$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9959 = 0.0041$

7 0

3 years ago

Find two consecutive integers whose sum is 93. (Show work)

kenny6666 [7]

You are looking for three consecutive integers whose sum is 93. First, divide 93 by 3 and get 31. Go one above 31 and one below. You get 30, 31 and 32 which sum to 93.

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3 years ago