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Snowcat [4.5K]
3 years ago
11

Write the prime factorization of each number in e-g. e. 25 f. 100g. 16​

Mathematics
1 answer:
kipiarov [429]3 years ago
8 0

Answers:

25 = 5*5 which is the same as 25 = 5^2

100 = 2*2*5*5 which is the same as 100 = 2^2*5^2

16 = 2*2*2*2 which is the same as 16 = 2^4

====================================================

Explanation:

To find the prime factorization of 25, we divide by prime numbers. We see that 25/2 = 12.5 isn't a whole number, and neither is 25/3 = 8.33 (approx). But 25/5 = 5 is a whole number. The result is a prime number so we can stop here. This shows that 25 = 5*5 = 5^2

------------

We'll do the same for 100 as well. Start with dividing over 2

100/2 = 50

50/2 = 25

we can't do 25/2 as mentioned earlier, but we can say 25/5 = 5

So basically 100 = 2*2*25 = 2*2*5*5 = 2^2*5^2

-------------

The same will happen with 16 as well. We divide by 2 until we arrive at a prime number result

16/2 = 8

8/2 = 4

4/2 = 2

We have three 2's as denominators and that last result 2 is included in the set of prime factors as well to say 16 = 2*2*2*2 = 2^4

-------------

The factor tree for each is shown below. For something like 100, there are a few ways to draw out the tree. The same applies for 16 as well.

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Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times
Elina [12.6K]

Answer:

The  probability is   P(A') =  0.485

Step-by-step explanation:

Let assume that the number of computer produced by factory C is  k = 1  

 So  From the  question we are told that

       The number produced by  factory A is  4k =  4

        The  number produced by factory B is  7k  = 7

        The  probability of defective computers from A is  P(A) =  0.04

        The  probability of defective computers from B is  P(B)  =  0.02

        The  probability of defective computers from C is P(C) =  0.03

Now the probability of factory A producing a defective computer out of the 4 computers produced is  

       P(a) =  4 *  P(A)

substituting values

        P(a) =  4 * 0.04

        P(a) = 0.16

The probability of factory B producing a defective computer out of the 7 computers produced is  

       P(b) = 7  *  P(B)

substituting values

        P(b) =  7 * 0.02

        P(b) = 0.14

The probability of factory C producing a defective computer out of the 1 computer produced is  

       P(c) = 1  *  P(C)

substituting values

        P(c) =  1 * 0.03

        P(b) = 0.03

So the probability that the a computer produced from the three factory will be defective is  

     P(t) =  P(a) +  P(b) +  P(c)

substituting values

     P(t) =   0.16  + 0.14 +  0.03

     P(t) =   0.33

Now the probability that the defective computer is produced from factory A is

      P(A') =  \frac{P(a)}{P(t)}

       P(A') =  \frac{ 0.16}{0.33}

        P(A') =  0.485

3 0
3 years ago
Given ΔRST : ΔLMN, which of the following is true<br> 1.
kvasek [131]
I hope this helps you



m (R)=m (L)


m (S)=m (M)


m (T)=m (N)
8 0
3 years ago
Read 2 more answers
A population of deer mice consists of 200 individuals and has an r = 0.1. A population of wood rats consists of 100 individuals
Mariulka [41]

Answer:

Population of dice mice after one year = 2000

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Step-by-step explanation:

The expression for population = dN/dt = rN

Upon integration N = rN²/2

Therefore for population N = 200 and r= 0.1

N after one year = (0.1 x 200²)/ 2 = 2000

Therefore for population N = 100 and r= 0.2

N after one year = (0.2 x 100²)/ 2 = 1000

Hence the population of deer mice is growing faster than the popular of wood rats

7 0
3 years ago
What’s equivalent to 25^x
Leviafan [203]

Answer:

5^2x

Step-by-step explanation:

write the number in exponential form with a base of 5

(5^2)^x

simplify the expression by multiplying exponents

5^2x

3 0
3 years ago
Please help! what equation best models this data?
kolezko [41]

Answer:

P = 309.35 + 2.31t

Step-by-step explanation:

The relation between the population of the USA and the time in years after 2010 will be linear as the increase in population is constant for every year since 2010 which is  2.31 million.

So, we can model the population P in million as a function of time(t) in years since 2010 as

P = 309.35 + 2.31t  ....... (1)

Now, for t = 0 i.e. in the year 2010, the population will be obtained from equation (1) to be 309.35 million.

(Answer)

8 0
3 years ago
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