All categories

3 years ago

Rabbits doubles every month. if you start with 2, how many will you have after 5 years

Mathematics

1 answer:

Illusion [34]3 years ago

6 0

2.30584301E+18 rabbits will be around bc you have to double the number 60 times

You might be interested in

Apparently I violated Community Guidelines ;-; AAAAAAAAAAAAAA

Archy [21]

Answer:

how? And I'm sorry :(

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is 75% of 24? Thanks!

FrozenT [24]

18 i looked it up hope its right

5 0

3 years ago

Read 2 more answers

If a2> 20, then which of the following could be a value of a?

PIT_PIT [208]

Answer:

Step-by-step explanation:

The square root of 20 = 4.47 so a for sure.

But the root of 20 is also -4.48 so that means that 2 and 3 are added to the list

Now what to do about - 8

(- 8)^2 = 64

The square root of 64 = +/- 8

It's kind of slippery but it depends on how you see the problem. It is a bit ambiguous.

I would eliminate - 8 but your instructor may not. Argue that it should not be included, but don't take the argument on forever. Let it go.

a for sure.

b and c I think so.

-8 is iffy.

Answer a

7 0

3 years ago

How many tens are in 6000 hundreds?

Flauer [41]

1 hundred = 10 tens

Multiply by 6000.

6000 hundreds = 60,000 tens

6 0

3 years ago

Read 2 more answers

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

Note that: 1 is subtracted because it is a probability without replacement

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

3 years ago