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timofeeve [1]
3 years ago
7

Can you guys help me please simplify: (4a^2b^3c^5)^3

Mathematics
1 answer:
KatRina [158]3 years ago
3 0

Answer:

64a^6b^9c^15

hope that helps

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The height of a certain banner is equal to one third of its length. If the banner is 5 ft tall, write the equation that can be u
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Answer:

L=3h

Step-by-step explanation:

Let height of a banner be h, then we know that this is \frac{1}{3} of the length L of the banner; so we have

h=\frac{1}{3} L .

Now we can rearrange this equation to solve for the length of the banner given it's height:

L=3h

So there we have it!

If the banner is 5 ft tall, then it's length will be:

L=3*5=15ft

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3 years ago
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Answer:

1

Step-by-step explanation:

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Can someone just pick a number from 1-4​
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When in europe, you are told that you are eating a steak weighing 140 grams. this is equivalent to how many ounces?
grandymaker [24]
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6 0
3 years ago
The physical plant at the main campus of a large state university receives daily requests to replace fluorescent light bulbs. Th
ankoles [38]

We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.

First of all, we will find z-score corresponding to 38 and 56.

z=\frac{x-\mu}{\sigma}

z=\frac{38-38}{6}=\frac{0}{6}=0

Now we will find z-score corresponding to 56.

z=\frac{56-38}{6}=\frac{18}{6}=3

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is -3\sigma\text{ to }3\sigma.

We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.

We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

\frac{99.7\%}{2}=49.85\%

Therefore, approximately 49.85\% of lightbulb replacement requests numbering between 38 and 56.

6 0
3 years ago
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