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kotykmax [81]
3 years ago
5

Think about the value of 6 in this number:

Mathematics
1 answer:
Whitepunk [10]3 years ago
6 0
The value of 6 in 465.09 divided by 100 is 0.6 which is 6 tenths.
Tod do this, you need to move the 6 twice to the right so its place value will change and divide by 100.
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(m^-4/3)^1/4 divided by m^0 n^3/4(m^-3/2 n^2)
postnew [5]
\frac{(m^{-\frac{4}{3}})^{\frac{1}{4}}}{m^{0}(m^{-\frac{3}{2}}n^{2})} = \frac{m^{-\frac{1}{3}}}{m^{-\frac{3}{2}}n^{2}} = \frac{m^{\frac{3}{2}}}{m^{\frac{1}{3}}n^{2}} = \frac{m^{\frac{7}{6}}}{n^{2}} = \frac{\sqrt[6]{m^{7}}}{n^{2}} = \frac{\sqrt[6]{m^{6} * m}}{n^{2}} = \frac{\sqrt[6]{m^{6}}\sqrt[6]{m}}{n^{2}} = \frac{m\sqrt[6]{m}}{n^{2}}
6 0
3 years ago
PLEASE HELP FAST NO LINKS!! <br> 3/2 (x-6)= -3
e-lub [12.9K]

Answer:

x= 4

hope this helps!

4 0
3 years ago
Read 2 more answers
Estimate the sum of 8.43 + 8.12 + 7.98.
In-s [12.5K]

Answer:

we estimate it and get

the answer is 25.53

7 0
2 years ago
What expression shows the relationship between the value of any term and n, its position in the sequence for the given sequence?
mafiozo [28]
Answer: -3n-2. 

Because for the first term, for example, plug in 1, you'll get -3(1)-2=-3-2=-5, which is the first one. Try it with all terms. 
8 0
3 years ago
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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