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bonufazy [111]
2 years ago
15

Use prime factorization for 600

Mathematics
2 answers:
Zigmanuir [339]2 years ago
8 0
2^3x3x5^2 because the guy above said so and I agree
Sophie [7]2 years ago
3 0

Answer: 2^3 x 3 x 5^2

Step-by-step explanation:

All we have to do is divide 600

alright

Chopping off the two zeros:

6x100

Factoring 6:

2x3x100

Factoring 100:

2x3x10x10:

Factoring both 10s:

2x3x2x5x2x5

2^3 x 3 x 5^2

Therefore, we have factored 600 into 2^3 x 3 x 5^2

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A sequence is shown.
marin [14]

Answer:

Step-by-step explanation:

The given sequence of numbers is increasing in geometric progression. The consecutive terms differ by a common ratio, r

Common ratio = 6/3 = 12/6 = 2

The formula for determining the nth term of a geometric progression is expressed as

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Where

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6 0
3 years ago
You have 6 reindeer, Prancer, Rudy, Balthazar, Quentin, Jebediah, and Lancer, and you want to have 3 fly your sleigh. You always
velikii [3]

Answer:

120 ways.

Step-by-step explanation:

We have been given that you have 6 reindeer, Prancer, Rudy, Balthazar, Quentin, Jebediah, and Lancer, and you want to have 3 fly your sleigh. You always have your reindeer fly in a single-file line.

We will use permutation formula to solve our given problem as:

_{r}^{n}\textrm{C}=\frac{n!}{(n-r)!}

_{3}^{6}\textrm{C}=\frac{6!}{(6-3)!}

_{3}^{6}\textrm{C}=\frac{6\cdot 5\cdot 4\cdot 3!}{3!}

_{3}^{6}\textrm{C}=6\cdot 5\cdot 4

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Therefore, you can arrange your reindeer in 120 different ways.

5 0
3 years ago
Read 2 more answers
The probability that a student is accepted to a prestigious college is 0.3. If 5 students from the same school apply, what is th
hodyreva [135]

Answer:0.849185

Step-by-step explanation:

Binomial probability formula we will use:

$P(x)=\frac{n !}{x !(n-x) !} p^{x} q^{n-x}$

where,

n=5$\\$\mathrm{P}$ (probability of success) $=0.3$\\q=1-p

The computation of the probability will be :

$P(x \leq 2 ; 5,0.3)=P(x=0 ; 5,0.3)+P(x=1 ; 5,0.3)+P(x=2 ; 5,0.3)+P(x=3 ; 5,0.3)$\\$=\left[\frac{5 !}{5 !(5-0) !}(0.3)^{0}(1-0.3)^{5-0}\right]+\left[\frac{5 !}{5 !(5-1) !}(0.3)^{1}(1-0.3)^{5-1}\right]$$+\left[\frac{5 !}{5 !(5-2) !}(0.3)^{2}(1-0.3)^{5-2}\right]$$++\left[\frac{5 !}{5 !(5-3) !}(0.3)^{3}(1-0.3)^{5-3}\right]$$++\left[\frac{5 !}{5 !(5-4) !}(0.3)^{4}(1-0.3)^{5-4}\right]$\\$=0.1681+0.3601+0.3087+0.006615+0.00567$\\$\Rightarrow 0.849185$

The probability that at most 4 are accepted = 0.849185

6 0
3 years ago
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