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2 years ago

How many times larger than 2x10^3 is 6x10^6?

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

3 0

Answer: 5,998,000

Step-by-step explanation:

2x10^3= 2,000

6x10^6= 6,000,000

In translation, Use 1 x The Initial SIG FIGS, or first numbers. 10x10 is 100, and notice how 10 and 10 have 2 zeros, so 1x1 = 1 +00 is a way to construct the number.

In this case, the 10^x, the x can represent the number of zeros after our first value.

So, 6 + six zeros, or 6,000,000

and, 2 + 3 zeros, or 2,000

To find the difference, subtract the numbers from eachother and extract the absolute value, (meaning the final answer won't be negative).

So, 6,000,000 - 2,000 = 5,998,000, which is the answer to the prompt!

Hope this Helped!

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Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

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steposvetlana [31]

<h2>PLEASE MARK BRAINLIEST!</h2>

Answer:

All these answers are correct. I triple - checked!

Step-by-step explanation:

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First question:

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3 years ago

Solve for x: 3x+2y=7

iVinArrow [24]

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3 years ago

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guajiro [1.7K]

The answer is A. 5/13

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3 years ago

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Alexxx [7]

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3 years ago

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