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Kipish [7]
3 years ago
11

ISO 9000 currently stresses _____ of a certified organization. A. inclusion of reused components in the office equipment B. mini

mizing harmful environmental effects C. a minimum of four supervisory levels D. product diversity E. continual improvement
Social Studies
1 answer:
VikaD [51]3 years ago
4 0

The correct answer is letter E.

Explanation: In ISO 9000 certification, Continual improvement is emphasized.

By reducing the variability in the product or service is an important key to quality. Poor quality has a positive effect on productivity because it usually takes longer to produce a good part.

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Answer: Yes I do think knowing more than one language is beneficial. I think this because if you know another language then you can talk with people who know the same language and also if you know another language then you could be smarter. That is why I think that knowing another language is beneficial for yourself and others.

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Where does the supreme court derrive its power in authorithy
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B) Article 3 of the US Constitution

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Read 2 more answers
Why did the professional dog walker go out of business math worksheet answers?
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Why Did the Professional Dog Walker ao Out of business, math worksheet Answers:

Q1. sin27°   = x/8

Solution:

We have to solve for x, therefore, we will rearrange the given equation for x.

We get,
x = 8 × sin27°

Using the calculator,

sin27° = 0.45

Now substitute the value of sin27° into the main equation.

we get,
x = 8 × 0.45
x = 3.63 (rounded to the nearest hundredth)


Q2. tan 18°  = n / 75

Solution:
We have to solve for n, therefore, we will rearrange the given equation for n.
We get,
n = 75 × tan 18°
Using the calculator,
tan 18° = 0.32
Now substitute the value of tan 18° into the main equation.
we get,
x = 75 × 0.32
x = 24.37 (rounded to the nearest hundredth)

Q3. sin40°  = 4 / a

Solution: We have to solve for a, therefore, we will rearrange the given equation for a.
We get,
a = 4 ÷ sin40°
Using the calculator,
sin40° = 0.64
Now substitute the value of sin40° into the main equation.
we get,
a = 4 ÷ 0.64
a = 6.25 (rounded to the nearest hundredth)

Q4. cos5°   = 92 / y

Solution: We have to solve for y, therefore, we will rearrange the given equation for y.
We get,
y = 92 ÷ cos5°
Using the calculator,
Cos5° = 0.99
Now substitute the value of cos5° into the main equation.
we get,
y = 92 ÷ 0.99
y = 92.92 (rounded to the nearest hundredth)

Q5:
Given the shape attached, therefore, using the triangle given, we have:
Angle of elevation = 35°
length of Opposite side to the angle = x
Length of Hypoteneus = 12
Calculations:
Using the SOH CAH TOA rules:
SOH stands for SineФ = Opposite ÷ Hypotenuse.

CAH stands for CosineФ = Adjacent ÷ Hypotenuse.

TOA stands for TangentФ = Opposite ÷ Adjacent.

Hence,

               SineФ = Opposite ÷ Hypotenuse

Substituting the values:

               Sine35° = x ÷ 12

               0.5735  = x ÷ 12

                          x = 0.5735 × 12

                          x = 6.88 (rounded to the nearest hundredth)

Q6: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 54°
length of the adjacent side to the angle = x
Length of Hypoteneus = 30
Calculations:
Using the SOH CAH TOA rules:

Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Cos54° = x ÷ 30

               0.5877  = x ÷ 30

                          x = 0.5877 × 30

                          x = 17.63 (rounded to the nearest hundredth)

Q7: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 22°
length of the adjacent side to the angle = 85
length of the opposite side to the angle = x
Calculations:

Using the SOH CAH TOA rules:

Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan22° = x ÷ 85

              0.4040 = x ÷ 85

                          x = 0.4040 × 85

                          x = 34.34 (rounded to the nearest hundredth)

Q8: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 16°
length of the opposite side to the angle = x
Length of Hypoteneus = 14
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Sine16° = x ÷ 14

               0.2756  = x ÷ 14

                          x = 0.2756 × 14

                          x = 3.86 (rounded to the nearest hundredth)

Q9: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 65°
length of the adjacent side to the angle = 9
length of the opposite side to the angle = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan65° = x ÷ 9

               2.1445 = x ÷ 9

                          x = 2.1445 × 9

                          x = 19.30 (rounded to the nearest hundredth)

Q10: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 51°
length of the adjacent side to the angle = x
Length of Hypoteneus = 70
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Cos51° = x ÷ 70

              0.6293  = x ÷ 70

                          x = 0.6293 × 70

                          x = 44.05 (rounded to the nearest hundredth)

Q11: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 36°
length of the opposite side to the angle = 15
Length of Hypoteneus = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Sine36° = 15 ÷ x

               0.5877  = 15 ÷ x

                          x = 15 ÷ 0.5877

                          x = 25.52 (rounded to the nearest hundredth)

Q12: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 60°
length of the adjacent side to the angle = x
length of the opposite side to the angle = 100

Calculations:
Using the SOH CAH TOA rules:

Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan65° = 100 ÷ x

               2.1445 = 100 ÷ x

                          x = 100 ÷ 2.1445

                          x = 46.63 (rounded to the nearest hundredth)

Q13: When a 25-ft ladder is leaned against a wall, it makes a 72° with the ground. How high up on wall does the ladder reach?

Solution: Given the shape attached, therefore, using the triangle given, we have:

The angle of elevation from the ground = 72°
length of the wall opposite to the angle = X
Length of ladder (Hypoteneus) = 25 feet

Calculations:
Using the SOH CAH TOA rules:
Hence,

               SineФ = Opposite ÷ Hypotenuse

Substituting the values:

               Sine72° = x ÷ 25

               0.9510  = x ÷ 25

                          x = 25 ÷ 0.9510

                          x = 23.77 (rounded to the nearest hundredth)


ANSWERS TO QUESTION 14 AND  15 ARE ATTACHED

7 0
3 years ago
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