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Effectus [21]
3 years ago
15

What is the prime factorization of the number 55?

Mathematics
2 answers:
Simora [160]3 years ago
6 0

Answer:

11 if im right :)

Step-by-step explanation:

LenaWriter [7]3 years ago
6 0

Answer:Thus, the prime factorization of 55 is 55 = 5 × 11. From the prime factorization of 55, it is clear that 5 and 11 are the prime factors of 55. We know that 1 is the factor of every number. Thus, the factors of 55 by prime factorization are 1, 5, 11, and 55.   Step-by-step explanation:

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Step-by-step explanation:

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What is the name of the property given below ? If ab = 0 then a = 0 , b = 0 or both a = 0 and b = 0 . A. Zero property B. Distri
FromTheMoon [43]

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c. zero product rule

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Maria finishes ¾ of the race in 6 hours. How long was the entire race?
GuDViN [60]

Answer:

We are asked to find the duration of entire race. As Rachel finished of the race in 6 hours, so of x will be equal to 6. Therefore, the duration of entire race was 8 hours.

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3 years ago
What is the value of x in the equation 150/84 = 25/x
Fofino [41]

Answer:

x = 14

Step-by-step explanation:

\frac{150}{84}  =  \frac{25}{x}  \\  \\ 150 \times x = 84 \times 25 \\  \\ x =  \frac{84 \times 25}{150}  \\  \\ x =  \frac{84}{6}  \\  \\ x = 14 \\

5 0
3 years ago
Read 2 more answers
A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th
agasfer [191]

Answer:

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

P(6) = 6×P(1)

P(T) = 21×P(1)

Where;

P(x) is the probability of getting number x on the dice.

P(T) is the total probability of obtaining any number

N(x) is the number of possible number x in terms of the distribution function.

P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

P(x) is directly proportional to N(x)

So, equation 1 becomes;

P(x) = N(x)/N(T) = P(x)/P(T) ....2

The probability of getting either a 5 or a 2 in one throw

P(2U5) = (P(2) + P(5))/P(T)

Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

P(2U5) = 7P(1)/21P(1)

P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

8 0
3 years ago
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