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kaheart [24]
3 years ago
9

I just need help on this problem please. Thanks :)

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0
All of them are the correct answer.
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A​ number, one-fifth of that​ number, and one-sixth of that number are added. the result is 82. what is the original​ number?
rusak2 [61]
X + .2x + .(1/6) x = 82
x = 60

5 0
3 years ago
Can u plz help me answer question 30
slamgirl [31]

We are given:

3x^3-2x^2

There seems to be nothing I can do to simplify this equation. However, I can factor out an x^2 since they both have at least an x^2. Here is what we get:

x^2(3x-2)

This is in the simplest form I can get into. I cannot do anything more to simplify this expression. This is your final answer.

3 0
3 years ago
Drag each expression to show whether it is equivalent to
Kazeer [188]

Answer:

We need to find which expressions are equivalent to 6(c+6), 6c+6 or neither.

6c+12: We extract the greatest common factor which is 6. Remember, when we extract a GCM, we divide each term by it.

6c+12=+(c+2)

Therefore, this expression is equivalent to neither of the given expressions.

2(3c+3): We just need to apply the distributive property.

2(3c+3)=6c+6

Therefore, this expression is equivalen to 6c+6.

We use the same process to the other expressions.

(6c)+(6\times 6)=6c+36=6(c+6)

3c+6+3c=6c+6

6c+36=6(c+6)

(6+c)+(6+6)=c+18, equivalent to neither.

7 0
3 years ago
F(x) = 5.22 – 3 on 2, 2+h]
Vika [28.1K]

Answer:

Step-by-step explanation:Wait im confused.

4 0
3 years ago
One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probab
Andrews [41]
You only need to consider the situations where 10 or 11 of the babies are girls, then subtract those probabilities from 1.  This will give probability that any other number up to 9 of the babies are girls.

Use binomial theorem.
P(x=k) = (nCk) p^k (1-p)^{n-k}

n = 11
k = 10,11
p = 1/2

P(x=10) = 11 (\frac{1}{2})^{11} = \frac{11}{2048} \\  \\ P(x=11) = 1(\frac{1}{2})^{11} = \frac{1}{2048} \\  \\ P(x \leq 9) = 1 - \frac{11}{2048} - \frac{1}{2048} \\  \\ P(x \leq 9)=\frac{2036}{2048} = \frac{509}{512}
4 0
3 years ago
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