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mafiozo [28]
2 years ago
9

Solve (1/8)^-3a=512^3a

Mathematics
2 answers:
lukranit [14]2 years ago
5 0
Solve (1/8)^-3a = 512^3a
A=0

I took the assignment it’s right.
melamori03 [73]2 years ago
5 0

Answer:

Using exponent rule:

\frac{1}{a^n}=a^{-n}

(a^n)^m = a^{nm}

Solve:

(\frac{1}{8})^{-3a} =(512)^{3a}

We can write 512 as:

512 = 8 \cdot 8 \cdot 8 = 8^3

then;

(\frac{1}{8})^{-3a} =(8^3)^{3a}

⇒(\frac{1}{8})^{-3a} =(8)^{9a}

Using exponent rule we have;

⇒(8^{-1})^{-3a} =(8)^{9a}

⇒(8)^{3a} =(8)^{9a}

On comparing both sides we have;

3a = 9a

⇒9a-3a = 0

⇒6a = 0

⇒a = 0

Therefore, the value of a = 0

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Answer:

The slope of the line is \displaystyle -\frac{1}{5}.

Step-by-step explanation:

We are given two coordinate points:

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  • (9, 1)

We are asked to find the slope of the line.

We can use the rise-over-run formula to solve for the slope of the line.

\displaystyle \text{slope} = \frac{\text{rise}}{\text{run}}\\\\\text{slope} = \frac{y_2-y_1}{x_2-x_1}

However, we firstly need to name our coordinate points.

In math, we can label our coordinates using the following label system:

(x_1, y_1), (x_2, y_2)

Therefore, we can also label our coordinates as such:

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  • x_2 = 9
  • y_2 = 1

Now, we can supply these values into the formula and solve for our slope, or a better known variable, <em>m</em>.

\displaystyle m = \frac{1 - 2}{9 - 4}\\\\m = \frac{-1}{5}\\\\m = -\frac{1}{5}

Therefore, our slope is \displaystyle -\frac{1}{5}.

8 0
3 years ago
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Archie is building a model of the Titanic. He is using a scale factor where 2 inches represents 50 feet. If the Titanic was 882.
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Suppose we have an unfair coin that its head is twice as likely to occur as its tail. a)If the coin is flipped 3 times, what is
Alenkinab [10]

Answer: a) 0.2222, b) 0.3292, c) 0.1111

Step-by-step explanation:

Since we have given that

Let the probability of getting head be p.

Since,  its head is twice as likely to occur as its tail.

p+\dfrac{p}{2}=1\\\\\dfrac{3p}{2}=1\\\\p=\dfrac{2}{3}

a)If the coin is flipped 3 times, what is the probability of getting exactly 1 head?

So, here, n  = 3

p=\dfrac{2}{3}

q=\dfrac{1}{3}

Now,

P(X=1)=^3C_1(\dfrac{2}{3})^1(\dfrac{1}{3})^2=0.2222

b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails?

2 tails means 3 heads.

So, it becomes,

P(X=3)=^5C_3(\dfrac{2}{3})^3(\dfrac{1}{3})^2=0.3292

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P(X\leq 1)=\sum _{x=0}^1^4C_x(\dfrac{2}{3}^x(\dfrac{1}{3})^{4-x}=0.1111

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