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Lina20 [59]
3 years ago
5

Hi! ❤️ , im looking for some help here. ill give brainliest if able to.

Mathematics
1 answer:
olchik [2.2K]3 years ago
5 0

Hello there! The solution of your question has been given below.

<h3>Solution:</h3>

{6}^{ - 2}  \\  = ( \frac{1}{6} ) ^{2}  \\  =  \frac{ {1}^{2} }{ {6}^{2} }  \\  =  \frac{1}{36}

<h3>Answer:</h3>

Option B.

\frac{1}{36}

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Simplify the answer -12 ÷ 3 × (-8 + (-4)² - 6) + 2 I need the steps for the equation on how to solve it.
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Answer:

-6

Step-by-step explanation:

We use PEMDAS to solve this,

so P stands for parentheses, so that's where we start.

We first, square the innermost parentheses with the exponent which is the E in PEMDAS, then then the outer parentheses

-12/3*(-8+16-6)+2

-12/3*(2)+2

Now we divide as in Division in PEMDAS.

-4*2+2

Now we multiply as in Multiplication in PEMDAS.

-8+2

Now we add as in A for Addition

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In PEMDAS, Multiplication doesn't always come before division, and same for addition and subtraction.

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3 years ago
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4 years ago
Joe owns a stock which has probability .5 of going up. This morning, he bought a ticket in a lottery game which gives him a prob
Anit [1.1K]

Answer:

The probability that Joe's stock will go up and he will win in the lottery  is 0.00005.

Step-by-step explanation:

Let the events be denoted as:

<em>X</em> = the stock goes up

<em>Y</em> = Joe wins the lottery

Given:

P (X) = 0.50

P (Y) = 0.0001

The events of the stock going up is not dependent on the the event of Joe winning the lottery.

So the events <em>X</em> and <em>Y</em> are independent of each other.

Independent events are those events that can occur together at the same time.

The joint probability of two independent events <em>A</em> and <em>B </em>is,

P(A\cap B)=P(A)\times P(B)

Compute the value of P (<em>X ∩ Y</em>) as follows:

P(X\cap Y)=P(X)\times P(Y)=0.50\times 0.0001=0.00005

Thus, the probability that Joe's stock will go up and he will win in the lottery  is 0.00005.

3 0
3 years ago
Read 2 more answers
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