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tresset_1 [31]
3 years ago
7

Solve the equation.

Mathematics
2 answers:
DENIUS [597]3 years ago
6 0
Hello :)

11,3x+7,2=86,3
⇔11,3x=86,3-7,2=79,1
⇔x=79,1/11,3= 7

The answers is c) 7 :). 

Oduvanchick [21]3 years ago
4 0

Answer:

The correct option is C) 7.

Step-by-step explanation:

Consider the provided equation.

11.3x+7.2=86.3

We need to solve the provided equation for x.

Subtract 7.2 from both side.  

11.3x+7.2-7.2=86.3-7.2

11.3x=79.1

Divide both sides by 11.3.

\frac{11.3x}{11.3}=\frac{79.1}{11.3}

x=7

Hence, the value of the equation is x=7.

Thus the correct option is C) 7.

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Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

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Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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