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3 years ago

A highway crossed a river that was 4.25 kilometers wide how wide is the river in decameters?

Mathematics

1 answer:

S_A_V [24]3 years ago

3 0

Answer:

425 decameters

Step-by-step explanation:

1 kilometer is equal to one decameter

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Create an expression equal to 7 with these numbers and symbols 1.3 (4.6 4.5) + ÷

Ber [7]

4.6+4.5=9.1÷1.3=7
so expression is (4.6+4.5)÷1.3=7

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3 years ago

David has a wooden board that is 6 feet long. How many pieces can be cut from the board if the length of each piece is 13 of a f

balandron [24]

Answer:

We are dividing a slab of wood into equal pieces, so the obvious function would be to divide. And how long are the pieces going to be? 1/4 of a foot. so that gives us:

6 ÷ ¼.

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6 * 4

Solve that, and you'll have your answer.

Step-by-step explanation:

Hope this answer helps you :)

Have a great day

Mark brainliest

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3 years ago

Find the first term and the common difference of the arithmetic sequence described. Give a recursive formula for the sequence. F

Rama09 [41]

Answer:

Step-by-step explanation:

375 7579 779

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4 years ago

What is the distance between -3 and 3 on a number line?<br> A. -6<br> B.6<br> C. 3<br> D. 0

SpyIntel [72]

The distance is 6

| | | | | | |
-3 -2 -1 0 1 2 3

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3 years ago

5.53 An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day. What

Licemer1 [7]

Answer:

a) 38.4% probability that on a given day this item is requested more than 5 times.

b) 0.67% probability that on a given day this item is not requested at all.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.

This means that $\mu = 5$

What is the probability that on a given day this item is requested

(a) more than 5 times?

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

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

We want P(X > 5). So

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

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

$P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337$

$P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842$

$P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404$

$P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755$

$P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 + 0.1755 = 0.616$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384$

38.4% probability that on a given day this item is requested more than 5 times.

(b) not at all?

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

0.67% probability that on a given day this item is not requested at all.

3 0

3 years ago