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Serhud [2]
3 years ago
5

Find the value of x A.)20 B.)55 C.)70 D.)38

Mathematics
2 answers:
lord [1]3 years ago
8 0

Answer:

Hello! answer: x = 20

Step-by-step explanation:

This is a complementary angle meaning it will add up to 90 degrees so...

20 × 4 = 80 80 - 10 = 70

70 + 20 = 90 therefore x = 20

Alex Ar [27]3 years ago
4 0

Answer:

A.) 20

Step-by-step explanation:

These angles are complementary meaning they add up to 90°.

4x - 10 + x = 90

5x - 10 = 90

5x = 100

x = 20

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Parallel to y = -x -5; through (4,-3)
eduard

Answer:

y = -x + 1

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer/how I got this answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

7 0
2 years ago
Read 2 more answers
A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li
hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote <em>R</em> as the event that a component needs rework.

It is given that:

P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20

Compute the probability that a randomly selected component needs rework as follows:

P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069}  =0.3623

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

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Answer:

<em>All real solutions.</em>

Step-by-step explanation:

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hope this halps... :D lol

4 0
3 years ago
Ten cards numbered 1 through 10 are mixed together and then one card is drawn. Find the probability of each event. Write the ans
slava [35]

Answer

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