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Mrac [35]
2 years ago
8

12 to the 3rd power divided by 12 to the 7th power ​

Mathematics
2 answers:
Molodets [167]2 years ago
7 0

Answer:

0.000048..............

ella [17]2 years ago
4 0

Answer:

0.00004822

Step-by-step explanation:

12^7= 35,831,808

12^3= 1728

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Find the greatest common divisor of 63 and 42
e-lub [12.9K]

Answer:

21

Step-by-step explanation:

7 0
3 years ago
in 1950, the U.S. federal budget was $39.4 billion, in 2000, the federal budget was $2025.2 billion. The exponential function is
Slav-nsk [51]

Answer:

The budget in 2010 is $4453.10915 billion

Step-by-step explanation:

We are given exponential function as

f(t)=39.4e^{0.0787932t}

where t is the time in years since 1950

now, we can find time in 2010

t=2010-1950=60

now, we can plug t=60

and we get

f(60)=39.4e^{0.0787932\times 60}

we get

f(60)=4453.10915

So, the budget in 2010 is $4453.10915 billion

3 0
3 years ago
The last school year enrollment of Genoa Middle school was 465 students. This year the enrollment is 525.
dimaraw [331]

Answer:

465 to 525

Step-by-step explanation:

60 difference

6 0
3 years ago
A card is drawn from a deck of 52 cards. match the correct fraction with its corresponding probability. 1. p(a diamond or a hear
vagabundo [1.1K]
1. p=26/52
26/52=.5
50 percent

2. p=29/52
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56 percent
5 0
3 years ago
Read 2 more answers
As part of quality-control program, 3 light bulbs from each bath of 100 are tested. In how many ways can this test batch be chos
hichkok12 [17]

Answer:

<h3>By 161700 ways this test batch can be chosen.</h3>

Step-by-step explanation:

We are given that total number of bulbs are = 100.

Number of bulbs are tested = 3.

Please note, when order it not important, we apply combination.

Choosing 3 bulbs out of 100 don't need any specific order.

Therefore, applying combination formula for choosing 3 bulbs out of 100 bulbs.

^nCr = \frac{n!}{(n-r)!r!} read as r out of n.

Plugging n=100 and r=3 in above formula, we get

^100C3 = \frac{100!}{(100-3)!3!}

Expanding 100! upto 97!, we get

=\frac{100\times 99\times 98\times 97!}{97!3!}

Crossing out common 97! from top and bottom, we get

=\frac{100\times 99\times 98}{3!}

Expanding 3!, we get

=\frac{100\times 99\times 98}{3\times 2\times 1}

= 100 × 33  × 49

= 161700 ways.

<h3>Therefore,  by 161700 ways this test batch can be chosen.</h3>
3 0
3 years ago
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