All categories

3 years ago

What is the least common multiple of 9 and 8

Mathematics

2 answers:

Cloud [144]3 years ago

8 0

Answer:

Frequently Asked Questions on LCM of 8 and 9

Find the prime factorization of 8 8 = 2 x 2 x 2

Find the prime factorization of 9 9 = 3 x 3

Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm: LCM = 72 = 2 x 2 x ...

Therefore, the least common multiple of 8 and 9 is 72.

Step-by-step explanation:

leonid [27]3 years ago

6 0

Answer:

72 is the least common multiple

You might be interested in

I need help. You can have brainliest. Plz <br> Thanks

Katena32 [7]

Answer:

Step-by-step explanation:

3 0

3 years ago

-3y+1/3=-5/6y-5/2 what does y equal

kotykmax [81]

Answer:

17/13

Step-by-step explanation:

$- 3y + \frac{1}{3} = - \frac{5}{6}y - \frac{5}{2} \\ - \frac{18}{6}y + \frac{2}{6} = - \frac{5}{6} y - \frac{15}{6} \\ \frac{2}{6} + \frac{15}{6} = - \frac{5}{6} y + \frac{18}{6} y \\ \frac{17}{6} = \frac{13}{6} y \\ \frac{6}{13} \times \frac{17}{6} = y \\ \frac{17}{13} = y$

8 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

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

<u>e)</u> $s = 621.76$

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

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

5 0

3 years ago

Plz help!

Marizza181 [45]

Answer:

800 is the answer

Step-by-step explanation:

90-40= 50---->4000/50=800

4 0

3 years ago

A cube has sides of length 5 cm.

worty [1.4K]

5 x 5 = 25
25 x 6 = 150

150cm^2

7 0

3 years ago

Read 2 more answers