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3 years ago

What is 28a^11 b^7 divided by 4a^3b

Mathematics

2 answers:

g100num [7]3 years ago

6 0

28 / 7 = 4 ;
a^11 / a^3 = a^(11 - 3 ) =a^8 ;
b^7 / b^1 = b^(7 - 1 ) = b^6 ;
The answer is 4a^8b^6

Veronika [31]3 years ago

4 0

Reduce the expression by cancelling the common factors.
7a^11b^7/a^3b

Reduce the expression by cancelling the common factors.
7a^8b^7/b

Reduce the expression by cancelling the common factors.
7a^8b^6

28a^11 b^7 divided by 4a^3b simplified is 7a^8b^6

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Answer:

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4 years ago

Herman is traveling 125 miles from his house to the beach by car. Herman plans to stop for lunch when the ratio of the distance

aliya0001 [1]

Answer:

250 miles

Step-by-step explanation:

The given distance between the house and the beach is 125 miles.

Let $x$ miles be the distance between the house and the lunch stop.

So, at the time of the lunch stop, he already traveled $x$ miles, the remaining distance is the distance between the lunch stop and the beach.

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3 years ago

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mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

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And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

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Step-by-step explanation:

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