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mr Goodwill [35]
3 years ago
14

Need help with this his one!

Mathematics
1 answer:
daser333 [38]3 years ago
7 0

Answer:

a) 6(6x-5)

=(6×6x) - (6×5)

=36x-30

b)11y+4y+10y

=y(11+4+10)

=y(25)

=25y

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1) Consider the first question. How many possible ways could you answer it?

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If you wrote out all the possibilities, how many combinations of answers would you get across the two questions?
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3/5 ÷ 2 as a Fraction
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1/3 AMOSC T.rhecc

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3 years ago
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−3x − 8y = 20 −5x + y = 19 solved by substitution
beks73 [17]

Delete as this is the wrong answer. Made a typo in the answer.

Answer:

x = \frac{-212}{3}

y = 24

Step-by-step explanation:

-3x - 8y = 20  .... (1)

-5x + y = 19   .... (2)

What we can do is take the (2) equation

-5x + y = 19

and solve for y by adding 5x to both sides

y = 24

now that we have y we substitute it in for y in equation (1)

-3x - 8(24) = 20

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we want to solve for x so we need to get its self.

add 192 to both sides

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now we divide both sides by -3. It's going to be a fraction.

x = \frac{-212}{3}

7 0
3 years ago
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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

c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails?

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

Hence, a) 0.2222, b) 0.3292, c) 0.1111

8 0
3 years ago
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