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2 years ago

14

How many distinct permutations can be made from the letters of the word columns? how many of these permutations begin with the l

etter m? assume you must use all of the letters?

1 answer:

luda_lava [24]2 years ago

6 0

7*6*5*4*3*2*1 = 7! = 5040

1*6*5*4*3*2*1 = 6! = 720

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Which linear equation represents on the graph

s2008m [1.1K]

Remember, rise/run is the easiest way to determine slope

y=mx+b is form

mx is slope, whilst b is y int

so, answer? y=-3x-2

Or A.

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2 years ago

A battery factory typically produces 12,000 batteries per day and uses 50 boxes for packing the batteries. If the number of boxe

weqwewe [10]

Answer:

25 more boxes is needed

Step-by-step explanation:

12000 batteries require 50 boxes. Let's find the unit rate.

That is, how many batteries are in each box??

That's 12,000/50!

12,000/50 = 240 batteries per box

Now, to find how many boxes we would need for 18,000 batteries, we will have to divide the total (18000) by 240(number of batteries per box). That is:

18,000/240 = 75 boxes

We would need 75 boxes to pack 18,000 batteries.

We want "how many MORE boxes needed", so we will the excess:

75 - 50 = 25 more boxes is needed

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1 year ago

If a plant grows 1/4 cm every day for 21 days, what is the total<br> amount of growth?

Naddika [18.5K]

I think it’s 5.25 cm but you might want to double check.

5 0

1 year ago

Read 2 more answers

$5000 was invested into an investment that pays 4.25% simple interest. The total value of the investment amounted to $6593.75. H

Basile [38]

Answer:

7.5 years

Step-by-step explanation:

P = $5000,

R = 4.25%,

A = $6593.75,

N =?

SI = A - P = 6593.75 - 5000 =$1593. 75

$\because \: SI = \frac{ PNR}{100} \\ \\ \therefore \: 1593.75 = \frac{5000 \times N \times 4.25}{100} \\ \\ \therefore \: 1593.75 \times 100 = 21,250 \times N \\ \\ N = \frac{159375}{21250} \\ \\ N = 7.5 \: years \:$

6 0

1 year ago

A karate center has 120 students. The director wants to set a goal to motivate her instructors to increase student enrollment. U

iVinArrow [24]

A) Plan A requires for a percentage increase of a number of students. This means that year after year the number of new students will increase. Plan B requires for a constant number of new students each year. This means that year after year the percentage increase would get smaller.

B) To solve this problem we will use formula for a growth of population:
$final = initial * (1+percentage)^{t}$
Where:
final = final number of students
initial = initial number of students
percentage = requested percentage increase
t = number of years

We can insert numbers and solve for t:
$240= 120* (1+0.12)^{t} \\ 2=1.12^{t} \\ log2=log(1.12^{t} ) \\ log2=t*log1.12 \\ t= \frac{log2}{log1.12} \\ t=6.12years$

For Plan B we can use simple formula
increase = 120
increase per year = 20
number of years = increase / (increase per year) = 120 / 20 = 6 years

Plan B is better to double the <span>enrollment.

C)We use same steps as in B) to solve this.
</span>

$360= 120* (1+0.12)^{t} \\ 3=1.12^{t} \\ log3=log(1.12^{t} ) \\ log3=t*log1.12 \\ t= \frac{log3}{log1.12} \\ t=9.69years$

For Plan B we can use simple formula
increase = 240
increase per year = 20
number of years = increase / (increase per year) = 240 / 20 = 12 years

Plan A is better to triple the enrollment.

3 0

1 year ago