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Igoryamba
3 years ago
13

Solve the equation.

Mathematics
2 answers:
Vika [28.1K]3 years ago
8 0

Hey there!!

Equation given :

... 78 = -2 ( m + 3 ) + m

... 78 = -2m - 6 + m

... 78 = -m - 6

... 78 + 6 = -m

... 84 = -m

... m = -84

Hope my answer helps!!

Thepotemich [5.8K]3 years ago
6 0

Hello there!

The answer should be the third option C.

78 = -2(m + 3) + m

First you had to simplify from both sides of the equations.

78=-2(m+3)+m

78=(-2)(m)+(-2)(3)+m

78=-2m+-6+m

78=(-2m+m)+(-6)

78=-m+-6

78=-m-6

Then you can flip by the equation.

-m-6=78

Add by 6 from both sides.

-m-6+6=78+6

-m=84

Divide by -1 from both sides.

\frac{-m}{-1}=\frac{84}{-1}

Simplify should be the correct answer.

m=-84

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

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Assume that 63% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below
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Answer:

a. 0.9931

b. 0.3423

c. 0.3907

d. 0.2670

e. 3.15

f. 1.0796

Step-by-step explanation:

The probability of the variable that said the number of left-handed people follow a Binomial distribution, so the probability is:

P(x)=nCx*p^{x}*(1-p)^{n-x}

nCx is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Where x is the number of left-handed people, n is the number of people selected at random and p is the probability that a person is left--handed. So P(x) is:

P(x)=5Cx*0.63^{x}*(1-0.63)^{5-x}

Then the probabilities P(0), P(1), P(2), P(3), P(4) and P(5) are:

P(0)=5C0*0.63^{0}*(1-0.63)^{5-0}=0.0069

P(1)=5C1*0.63^{1}*(1-0.63)^{5-1}=0.0590

P(2)=5C2*0.63^{2}*(1-0.63)^{5-2}=0.2011

P(3)=5C3*0.63^{3}*(1-0.63)^{5-3}=0.3423

P(x)=5C4*0.63^{4}*(1-0.63)^{5-4}=0.2914

P(x)=5C5*0.63^{5}*(1-0.63)^{5-5}=0.0993

Then, the probability P(x≥1) that there are some lefties among the 5 people is:

P(x≥1) = P(1) + P(2) + P(3) + P(4) + P(5)

P(x≥1) = 0.0590 + 0.2011 + 0.3423 + 0.2914 + 0.0993 = 0.9931

The probability P(3) that there are exactly 3 lefties in the group is:

P(3) = 0.3423

The probability P(x≥4) that there are at least 4 lefties in the group is:

P(x≥4) = P(4) + P(5) = 0.2914 + 0.0993 = 0.3907

The probability P(x≤2) that there are no more than 2 lefties in the group is:

P(x≤2) = P(0) + P(1) + P(2) = 0.0069 + 0.0590 + 0.2011 = 0.2670

On the other hand, the expected value E(x) and standard deviation S(x) of the variable that follows a binomial distribution is:

E(x)=np=5(0.63)=3.15\\S(x)=\sqrt{np(1-p)}=\sqrt{5(0.63)(1-0.63)}=1.0796

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