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3 years ago

Summary

1 answer:

5 0

It the multiplication property because you are multiplying 2x=20 so then you divide on both sides with 2 and that gives you 10

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3 years ago

Determine whether 33 divides the product of 2⋅3⋅7⋅11. If it does divide the product, find the quotient. If it does not divide th

swat32

Answer:

The answer would be 30

Step-by-step explanation

For positive integers: 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 = 2310 For all integers: ± ( 2 ⋅ 3 ⋅ 5 ) = ± 30 For Gaussian integers: ± 1 ± 3 i and ± 3 ± i (all combinations of signs) Explanation: A prime number is a number whose only factors are itself, units and unit multiples of itself. So in the positive integers, the first few primes are: 2 , 3 , 5 , 7 , 11 , ... So the smallest composite positive integer with the five smallest prime positive integers as factors is: 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 = 2310 If we extend our interest to include negative integers, then the smallest primes are: 2 , − 2 , 3 , − 3 , 5 , − 5 , ... So the smallest composite integers with the five smallest prime integers as factors are: ± ( 2 ⋅ 3 ⋅ 5 ) = ± 30 If we consider Gaussian integers, then the smallest primes are: 1 + i , 1 − i , − 1 + i , − 1 − i , 1 + 2 i , 1 − 2 i , − 1 + 2 i , − 1 − 2 i , 2 + i , 2 − i , − 2 + i , − 2 − i , 3 , − 3 ,... So the smallest composite Gaussian integers with the five smallest prime Gaussian integers as factor are: ( 1 + i ) ( 1 + 2 i ) = − 1 + 3 i , 1 + 3 i , − 1 − 3 i , − 1 + 3 i , 3 + i , 3 − i , − 3 + i , − 3 − i

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3 years ago

A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod

tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1 2 3

495 525 470

500 515 480

505 505 460

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4= 470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using μ ± σ , μ±2 σ or μ ± 3 σ.

In this question the limits are determined by using μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4= 470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other two means are within the given UCL and LCL.

This can be shown by the diagram.

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2 years ago