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3 years ago

What is 1/56 of 12769800000

Mathematics

2 answers:

atroni [7]3 years ago

7 0

228032142.857
Hope this helps

fiasKO [112]3 years ago

6 0

228032142.857 . i think it is the correct answer .

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How do i do distributive property

trasher [3.6K]

Expand the equation.
Multiply (distribute) the first numbers of each set, outer numbers of each set, inner numbers of each set, and the last numbers of each set.
Combine like terms.
Solve the equation and simplify, if needed.

7 0

2 years ago

Read 2 more answers

A pair of fair dice is cast. Let Edenote the event that the number landing uppermost on the first die is a 1 and let Fdenote the

iris [78.8K]

Answer:

They are not independent

Step-by-step explanation:

Given

E = Occurrence of 1 on first die

F = Sum of the uppermost occurrence in both die is 5

Required

Are E and F independent

First, we need to list the sample space of a roll of a die

$Event\ 1 = \{1,2,3,4,5,6\}$

Next, we list out the sample space of F

$Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\$

$6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}$

In (1): the sample space of E is:

$E = \{1\}$

So:

$P(E) = \frac{n(E)}{n(Event\ 1)}$

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

In (2): the sample space of F is:

$F = \{5,5,5,5\}$

So:

$P(F) = \frac{n(F)}{n(Event\ 2)}$

$P(F) =\frac{4}{36}$

$P(F) =\frac{1}{9}$

For E and F to be independent:

$P(E\ and\ F) = P(E) * P(F)$

Substitute values for P(E) and P(F)

This gives:

$P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}$

$P(E\ and\ F) = \frac{1}{54}$

However, the actual value of P(E and F) is 0.

This is so because $E = \{1\}$ and $F = \{5,5,5,5\}$ have 0 common elements:

So:

$P(E\ and\ F) = 0$

Compare $P(E\ and\ F) = \frac{1}{54}$ and $P(E\ and\ F) = 0$ .

These values are not equal.

Hence: the two events are not independent

6 0

3 years ago

The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a

Anettt [7]

Answer:

1) The size of the colony after 4 days is 6553 mosquitoes

2) After $t = 4.9\ days$

Step-by-step explanation:

To answer this question you must use the growth formula

$N = N_0e ^ {kt}$

Where

$N_0$ is the initial population of mosquitoes = 1000

t is the time in days

k is the growth rate

N is the population according to the number of days

We know that when t = 1 and $N_0=1000$ then N = 1600

Then we use these values to find k.

$1600 = 1000e ^{k(1)}\\\\\frac{1600}{1000} = e ^ k\\\\ln(\frac{1600}{1000}) = k\\\\k = 0.47$

Now that we know k we can find the size of the colony after 4 days.

$N = 1000e ^ {0.47(4)}\\\\N = 6553\ mosquitoes$

To know how long it should take for the population to reach 10,000 mosquitoes we must do N = 10000 and solve for t.

$10000 = 1000e ^ {0.47t}\\\\10 = e ^ {0.47t}\\\\ln(10) = 0.47t\\\\t = \frac{ln(10)}{0.47}\\\\t = 4.9\ days$

6 0

2 years ago

Many of you will live to the year 2100, which is not a leap year. Not all years which are divisible by 4 are leap years. A cente

Slav-nsk [51]

Answer:

For example, the years 1600, 2000, and 2400 are century leap years since those numbers are divisible by 400, while 1700, 1800, 1900, 2100, 2200, and 2300 are common years despite being divisible by 4.

Step-by-step explanation:

3 0

2 years ago

NEED HELP ASAP WILL GIVE BRAINLIEST REAL ANSWERS ONLY PLZ

Elis [28]

Answer:

Step-by-step explanation:

They are trying to see how long it takes people to complete the writing portion of the test.

4 0

2 years ago