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pychu [463]
3 years ago
13

The greatest common factor of 3m2n + 12mn2 is

Mathematics
2 answers:
Kamila [148]3 years ago
8 0
3m^2n+12mn^2
In both the sets, 3 and m as well as n can be found as common factors.
So,
3mn( m+4n)


Therefore the answer is 3mn (#3)
Elena L [17]3 years ago
6 0
3m²n + 12mn²
3mn(m) + 3mn(4n)
3mn(m + 4n)

The GCF of 3m²n + 12mn² is 3mn.
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Ryan is making pancakes for the Drama Club’s pancake breakfast. The table below shows the amount of pancake mix and milk needed
adell [148]
There is no table, and what us the question?
4 0
3 years ago
Help please whats the answer
Ad libitum [116K]
<h3>Answer:</h3>

12 students

<h3>Explanation:</h3>

40% earned a B; 20% earned an A, so (40%+20% =) 60% got a "B or better."

60% of 20 students is ...

... 60/100 × 20 students = 1200/100 students = 12 students

5 0
3 years ago
According to the Center for Disease Control and Prevention (CDC), up to 20% of Americans contract the influenza virus each year,
12345 [234]

Answer:

(1) a. 0.0009

(2) d. 0.640

(3)

  • a. P(A and B) = 0.06.
  • b. P(A or B) = 0.70.

(4)Not disjoint

(5) a. nearly 0.

(6)b. 0.919

Step-by-Step Explanation:

(1)Probability of a baby being born with a birth defect =3%=0.03

The probability that both babies have birth defects=0.03 X 0.03= 0.0009.

(2)The probability of contracting the influenza virus each year = 20%=0.2

Therefore, the probability of not contracting the influenza virus =1-0.2=0.8

The probability that neither baby catches the flu in a given year:

=0.8 X  0.8

=0.64

(3)

P(A)=0.1

P(B)=0.6

P(A or B)=P(A)+P(B)=0.1 + 0.6 =0.7

P(A and B)=P(A)XP(B)=0.1 X 0.6 =0.06

(4)

P(A)=0.2

P(B)=0.9

Event A and B cannot be disjoint.

(5)

The probability of an American woman aged 20 to 24 having Chlamydia infection  =\dfrac{2791.5}{100000}

The probability that three randomly selected women in this age group have the infection

=\dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \\\\=0.00002175\\\approx 0

(6)The probability of an American woman aged 20 to 24 not having Chlamydia infection  =1-\dfrac{2791.5}{100000}

The probability that three randomly selected women in this age group do not have the infection

=\left(1-\dfrac{2791.5}{100000}\right)^3\\\\=0.9186\\\approx 0.919

7 0
4 years ago
You pick a card, roll a die, and find the sum. How many different sums are possible?
svlad2 [7]

Answer:

6

Step-by-step explanation:

7 0
3 years ago
The graph below shows the distance, y, in feet, of a mouse from its hole, for a certain amount of time, x, in minutes:
Kobotan [32]
Based on the graph, what is the initial value of the graph and what does it represent <span>0.23 foot; it represents the original distance of the mouse from its hole</span>
5 0
4 years ago
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