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iren2701 [21]
2 years ago
9

36x – 8y2 when x = 3 and y = –6

Mathematics
2 answers:
sergiy2304 [10]2 years ago
8 0

Welcome to Brainly! :)

I cannot understand your equation, therefore I have put it as 36x + 8y + 2. If I'm wrong let me know and I'll fix everything. :)

We need to plug in our values for what we are given.

Let us learn about plugging in.

Let's start with an example equation:

1x + 3y + 1 - Values: y = 2, x = 4 This is a simple equation.

- When it tells you to plug in or tell you the values, that's your ket to plug in the variable for what it tells you.

Now, let's solve this example, our equation is:

1(4) + 3(2) + 1

Now, we have the original equation but plugged in. Next we simplify:

4 + 6 + 1

Add it up and get 11.

Great! Now that we are done with notes, we can solve.

36x – 8y + 2

It tells us that x = 3 an y = 6. Like the equation, we need to plug it in.

36(3) - 8(-6)+2

Great, now we need to simplify.

36(3)-3(-6)+2 \\ 108 + 18+2\\

Cool, now we simplify and get 128!

Great job!

I hope you had amazing experience with brainly! :D

OlgaM077 [116]2 years ago
6 0
This . (Dot) means multiply
36 - 8.-6.2
108 - (-96)
108 + 96
206
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= ABD + DBC

= 33.3° + 30.6° 

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b) If E is drawn directly opposite C.

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ABE + ABD + DBC = 180°

ABE + 33.3° + 30.6° = 180°

ABE + 63.9° = 180°

ABE = 180 - 63.1

ABE = 116.9° 

Hope this explains it.
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2 years ago
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Step-by-step explanation:

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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random
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The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
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In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is \frac{1}{3}

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

Therefore, P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}


Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}

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Answer:

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90,761 - 50,756 = 40,005

Over 30 years:

30*40,005 = 1,200,150

The difference in earnings, over a 30-year career, for men vs women, is $1,200,150

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