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lutik1710 [3]
2 years ago
7

Suppose $n$ is a positive integer such that $6n$ has exactly 9 positive divisors. How many prime numbers are divisors of $6n$

Mathematics
1 answer:
zvonat [6]2 years ago
6 0

The number of prime divisors of "6n" is 2 numbers 2 and 3

Given

"n"  = 6    

thus 6n  = 6(6) = 36

The positive divisors are   1, 2, 3, 4, 6, 9, 12, 18 and 36

So, the number of prime divisors of "6n" is 2 numbers 2 and 3.

The prime factors of a positive integer are the top numbers that divide that integer exactly. The technique of finding those numbers is known as integer factorization or prime factorization.

The fundamental theorem of mathematics says that every positive integer has a completely unique prime factorization.

Learn more about prime factorization here: brainly.com/question/1081523

#SPJ4

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2 years ago
Can get some help on 2,4,22, 42 and 50 please
Fittoniya [83]

Answer:

2. zeros

4. difference

22. 79,158.5

42. 0.0099

50. 4.37

Step-by-step explanation:

2. Only adding 0s after a decimal point does not change the value.

4. Since you are subtracting then you estimate difference.

22. Add each number using place value.

75285

        2

  3971

        0.5

__________

79258.5

42. 0.01 is really 0.0100. If you subtract 0.0001 then it is like subtracting 100-1 = 99 but in the decimal place. So it becomes 0.0099.

50. Subtract by lining up decimal places. Then move place values to subtract.

      9.00

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6 0
3 years ago
The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen
e-lub [12.9K]

Answer with Step-by-step explanation:

Since the given event is binary we can use Bernoulli's probability to sove the problem

Thus for an event 'E' with probability of success 'p' the probability that the event occurs 'r' times in 'n' trails is given by

P(E)=\frac{n!}{(n-r)!\cdot r!}\cdot p^{r}\cdot (1-p)^{n-r}

Part a)

For part a n = 11 , r =9, p = 0.75

Applying values we get

P(E)=\frac{11!}{(11-9)!\cdot 9!}\cdot (0.75)^{9}\cdot (1-0.75)^{11-9}\\\\\therefore P(E)=0.2581

Part b)

For part b n = 20 , r = 16 , p=0.75

Applying values we get

P(E)=\frac{20!}{(20-16)!\cdot 16!}\cdot (0.75)^{16}\cdot (1-0.75)^{20-16}\\\\\therefore P(E)=0.1896

3 0
3 years ago
3/9=21/k what is the strategy for solving it
Vinvika [58]

Answer: k = 63

Step-by-step explanation: You need to multiply the left side by its reciprocal to move the numbers to the right. You also need to multiply by k on the right side to get it alone on the left. This will result in

(9/3)(3/9) = (21/k)(9/3) simplified

1 = 189/3k • k    then multiply by k

k = 189/3 simplify

k = 63

5 0
3 years ago
What is equavalent 3 11/40.
Lyrx [107]

Answer:

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Step-by-step explanation:

<u>Estimate the fraction part of the number.</u>

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We can estimate the fraction part as 1/2 since its greater than 1/4.

The number becomes 3 1/2.

Best choice is C

4 0
3 years ago
Read 2 more answers
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