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2 years ago

What LCM of 9 and 21

Mathematics

2 answers:

jeyben [28]2 years ago

5 0

Answer:

Step-by-step explanation:

21: 21, 42, 63

9: 9, 18, 27, 36, 45, 54, 63

I am joyous to assist you anytime.

ratelena [41]2 years ago

4 0

The least common multiple of 9 and 21 is 63

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How do yo work out 9.1 divided by 3575

EastWind [94]

9.1/3575
use long division like this
 _0._________
3575|91.000000

we find how many of 3575 fit into the 91
it's too big so we go farther
how many 3575 go into 910
too big so we go farther
how many 3575 go into 9100
the answer is 2 so put that in the correct place and mulitply that and put that in the correct place. then we subtract

 _0.02_________
3575|91.000000
 -71.50
 19.50
bring down the next number
find how many go into 19.500
the answe ris 5

 _0.025_________
3575|91.000000
 -71.50
 19.500
 -17.875
 1.625

bring down he next zero ( I fast forward and skip steps for convinience)

 __
 _0.025455_________
3575|91.000000
 -71.50
 19.500
 -17.875
 1.6250
 -1.4300
 .19500
 -.17875
 16250
 -14300
and so on untill infinity so the answe ris 0.0254555555555555 (enless fives)

8 0

3 years ago

Can help me for homework pleas

Rasek [7]

Hey You!

32 / 8 = 4

4 × 7 = 28

20. ANSWER:

The fish swims 28 feet deep.

21. ANSWER:

Numerator;Denominator

22. ANSWER:

439 / 3 = 146.3333 repeat.

So, each individual computer costs more than each individual printer.

6 0

3 years ago

Read 2 more answers

40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability

aleksley [76]

Answer:

$(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4 = 0.6

Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.

P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

= $\binom{n}{0}p^{0}q^{n-0}$ + $\binom{n}{1}p^{1}q^{n-1}$ + $\binom{n}{2}p^{2}q^{n-2}$ + $\binom{n}{3}p^{3}q^{n-3}$

= $q^{n}+npq^{n-1}$ + $\frac{n(n-1)(p)^{2}q^{n-2}}{2}$ + $\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}$

= $(0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

7 0

3 years ago

Read 2 more answers

The following histogram presents the amounts of silver (in parts per million) found in a sample of rocks. One rectangle from the

gladu [14]

Answer:

The height of the missing rectangle is 0.15

Explanation:

The image attached has the mentioned histogram.

Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.

Only the rectangle for the class [3,4] is missing.

The height of each rectangle is the relative frequency of the corresponding class.

The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.

In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.

1. Sum of the known relative frequencies:

0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.85

2. Missing frequency:

1 - 0.85 = 0.15

3. Conclusion:

The height of the missing rectangle is 0.15

4 0

2 years ago

Jan bought 2 novels regularly priced at $24 and $18. Then reduced by 25%. What did she save?

max2010maxim [7]

Answer:

yeeeeeeeeeeeet

Step-by-step explanation:

thx for points

3 0

3 years ago