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3 years ago

Can anybody help me please I'm stuck with this question.

Mathematics

1 answer:

tekilochka [14]3 years ago

8 0

Answer: 18 units

Step-by-step explanation:since its is a horizontal line x2 - x1

7 -1 = 6

line 2:

since this is a vertical line y2 - y1

7 - 3 = 4

line 3:

since this is a horizontal line x2 - x1

7 - 4 = 3

line 4:

for this we need to use the distance formula which allows us to find the distance making a third point to form a right angle triangle

point 1: (1,3)

point 2: (4,7)

point 3 (new point) : (4,3)

now we can apply the pythogorean thereum (C squared = B squared + A squared) with the following lines.

line 1: (1,3) - (4,7)

line 2: (1,3) - (4,3)

line 3: (4,3) - (4,7)

line 1 squared = line2 squared + line 3 squared

calculate length of line 2 and 3

line 1 squared = (4 - 1) squared + (7 - 3) squared

line 1 squared = 3 squared + 4 squared

line 1 squared = 9 + 16

line 1 squared = 25

root both sides

line 1 = 5

add all the liens together

6 + 4 + 3 + 5 = 18

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Answer:

The solution is given below:

Step-by-step explanation:

The computation is shown below:

= 19 + 7 divided by 2 - 5

= (19 + 7) ÷ (2) - 5

= 26 ÷ 2 - 5

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Hence, after solving this the value would be 8

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2 years ago

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Answer:

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Step-by-step explanation:

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

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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

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

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

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

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

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I'm sorry if I misunderstood.

Good luck mate! :)

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3 years ago

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