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Snezhnost [94]
3 years ago
12

What is the probability of rolling a number greater than 2 with a standard number cube

Mathematics
1 answer:
disa [49]3 years ago
6 0

Answer:

4/6 or 2/3

Step-by-step explanation:

6 sides with 4 of the numbers being greater than 2; 4/6. The simplified version is 2/3.

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The length of the sides of a triangle is given. Determine
aalyn [17]

Answer:

the triangle is right

Step-by-step explanation:

The Pythagorean theorem dictates that a^2 +b^2=c^2

6^2+8^2=10^2

36+64=100

100=100

6 0
3 years ago
Two students were given the expression shown to simplify.
navik [9.2K]
<h3>The equivalent expression is 6 + 6x</h3>

<em><u>Solution:</u></em>

<em><u>Given expression is:</u></em>

9 - (3 - 6x)

We have to find the equivalent expression

Use the distributive property,

a(b + c) = ab + bc

Therefore, from given expression,

9 - 1(3 - 6x) = 9 - 3 + 6x

Combine the constants, 9 - 3 = 6

9 - 1(3 - 6x) = 6 + 6x

Thus the equivalent expression is 6 + 6x

3 0
3 years ago
Please help me please someone will help you please Help me
CaHeK987 [17]

Answer:

<h2>Sorry I couldn't find it.</h2>

Step-by-step explanation:

<h3>:(((((((((((((((((((</h3>
5 0
3 years ago
If a cyclist travels at a speed of 36km/h. what is the speed in minutes<br>​
Snowcat [4.5K]

Answer:

0.6 km / min

0.6 kilometres per minute

Step-by-step explanation:

Review the images and don't forget tu support me please!!! ;)

STEP BY STEP:

36 (km/h) (1h / 60min) = 0.6 km/min

8 0
2 years ago
Organisms A and B start out with the same population size.
Triss [41]

Answer:

At the end of ten days, the size of population B is 256 times that of population A

Step-by-step explanation:

We work under the premise that population A and B start both with the same number of individuals. Let's call such initial population N_0

Now, we write the exponential expression that describes population A as a function of days (t) for the first 6 days:

N_A=N_0\,(2)^t

which represents the starting point with N_0 individuals on day zero, doubling after one day (t= 1), and keeping on doubling the following days for 6 days.

So at the end of 6 days, population A would have the following number of individuals:

N_A=N_0\,(2)^6\\N_A=N_0\,(64)\\N_A=64\,N_0

That is 64 times the starting number of individuals.

After this, the population stops growing and starts reducing to one-half each day. This behavior can be represented by:

N_A=64\,N_0\,(\frac{1}{2} )^t

therefore after 4 days in this pattern, this culture has the following number of organisms:

N_A=64\,N_0\,(\frac{1}{2} )^4\\N_A=64\,N_0\,(\frac{1}{16} )\\N_A=4\,N_0

which is now just four times what the culture started with.

Now, on the other hand, population B grows doubling each day without interruption, so at the end of 10 days its size is given by:

N_B=N_0\,(2)^t\\N_B=N_0\,(2)^10\\N_B=N_0\1024\\N_B=1024\,N_0

that is it has 1024 times the initial number of organisms.

So if we compare both populations at day 10:

\frac{N_B}{N_A} =\frac{1024\/N_0}{4\,N_0} =256

Therefore, at the end of ten days, population B is 256 times the size of population A

4 0
3 years ago
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