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Anna35 [415]
4 years ago
11

What ratio can be used to calculate the geometric mean of 6 and 42?

Mathematics
1 answer:
Aleksandr [31]4 years ago
3 0
The ratio...i think they answer is idk
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What does not represent the quotient of 17 divided by 10?
JulijaS [17]

Answer:

C

Step-by-step explanation:

The quotient of 17 divided by 10 is the fraction 17/10 or simplified 1 7/10, 1.7 or 1 R 7. It is NOT 10 / 17. C is the correct answer since it does not represent the quotient described.

6 0
3 years ago
Find the product. (a-6)^2=
gavmur [86]

Answer:

a^2-12a+36

Step-by-step explanation:

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What is 4.735 rounded to the nearest one
JulsSmile [24]
So based on place values, the One's place is occupied by 4. To round to this digit, you go one number to the right. If that number is 5 or higher, you add one to 4. If it is less than 5, you don't add any to 4. Since 7 is greater than 5, you add 1 to 4, giving you 5, since you can now remove all the other numbers to the right.

5.


6 0
3 years ago
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The function G(x)=f(x+3)+2 which of the following is true of the graph of g(x)?
zepelin [54]
G(x)=(x+3)+2

G(x) = (x-h) + k
h translates the graph left/right and k translates the graph up/down.
Because the equation is x- the parenthesis is (x - - 3) which makes h a negative 3 and will translate left. The k positive so it will move up.
Letter B
7 0
3 years ago
A jar contains 6 blue, 4 red, and 2 green marbles. What is the probability that three blue marbles are drawn with out replacemen
ch4aika [34]

Answer:  The required probability is 9.09%.

Step-by-step explanation:  Given that a jar contains 6 blue, 4 red, and 2 green marbles.

We are to find the probability that three blue marbles are drawn with out replacement.

Total number of marbles = 6 + 4 + 2 = 12.

After drawing one blue marble, we are left with total 11 marbles. And after drawing the second blue marble, we are left with total 10 marbles.

Therefore, the probability of drawing three blue marbles without replacement is given by

p\\\\\\=\dfrac{^6C_1}{^{12}C_1}\times\dfrac{5C_1}{^{11}C_1}\times\dfrac{^4C_1}{^{10}C_1}\\\\\\=\dfrac{6}{12}\times\dfrac{5}{11}\times\dfrac{4}{10}\\\\\\=\dfrac{1}{11}\\\\\\=\dfrac{1}{11}\times100\%\\\\=9.09\%.

Thus, the required probability is 9.09%.

6 0
4 years ago
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