Rita's answer in incorrect because the remainder that Rita got which is 23 is greater 17 which is the divisor. 17 can still go in 23 1time to have a remainder of 6. Hence Rita's answer is wrong with 34R6 being the correct answer.

Step-by-step explanation:

Quotient is the required answer if 584 is divided by 17

584 will be the dividend and 17 is the divisor.

To calculate 584/17, 17 will go in 58 3times with 7 as remainder. This remainder 7 will combine with 4 to make 74, 17 will also go in 74 4times with remainder of 6.

The final answer by combining the number of times 17 goes in 584 which is 34 with final remainder of 6. This means that;

584/17 = 34 remainder 6 i.e 34R6

This shows that Rita's answer in incorrect because the remainder that Rita got which is 23 is greater 17 which is the divisor. 17 can still go in 23 1time to have a remainder of 6. Hence Rita's answer is wrong with 34R6 being the correct answer.

5 0

4 years ago

Suppose that Motorola uses the normal distribution to determine the probability of defects and the number of defects in a partic

Mekhanik [1.2K]

Answer:

$\mathbf{P(X < 9.85 \ or \ X> 10.15) \approx 0.3171}$ to four decimal places.

$\mathbf{P(X < 9.85 \ or \ X> 10.15) =0.0027}$ to four decimal places.

Step-by-step explanation:

Assuming X to be the random variable which replace the amount of defectives and follows standard normal distribution whose mean (μ) is 10 ounces and standard deviation (σ) is 0.15

The values of the random variable differ from mean by ± 1 \such that the values are either greater than (10+ 0.15) or less than (10-0.15)

= 10.15 or 9.85.

The probability that the amount of defectives which are either greater than 10.15 or less than 9.85 can be calculated as follows:

$P(X < 9.85 \ or \ X> 10.15) = 1-P ( \dfrac{9.85-10}{0.15}< \dfrac{X-10}{0.15}< \dfrac{10.15-10}{0.15})$

$P(X < 9.85 \ or \ X> 10.15) = 1- \phi (1) - \phi (-1)$

Using the Excel Formula ( = NORMDIST (1) ) to calculate for the value of z =1 and -1 ;we have: 0.841345 and 0.158655 respectively

$P(X < 9.85 \ or \ X> 10.15) = 1- (0.841345-0.158655)$

$P(X < 9.85 \ or \ X> 10.15) =0.31731$

$\mathbf{P(X < 9.85 \ or \ X> 10.15) \approx 0.3171}$ to four decimal places.

b) Through process design improvements, the process standard deviation can be reduced to 0.05.

The probability that the amount of defectives which are either greater than 10.15 or less than 9.85 can be calculated as follows:

$P(X < 9.85 \ or \ X> 10.15) = 1-P ( \dfrac{9.85-10}{0.05}< \dfrac{X-10}{0.05}< \dfrac{10.15-10}{0.05})$

$P(X < 9.85 \ or \ X> 10.15) = 1- \phi (3) - \phi (-3)$

Using the Excel Formula ( = NORMDIST (3) ) to calculate for the value of z =3 and -3 ;we have: 0.99865 and 0.00135 respectively

$P(X < 9.85 \ or \ X> 10.15) = 1- (0.99865-0.00135)$

$\mathbf{P(X < 9.85 \ or \ X> 10.15) =0.0027}$ to four decimal places.

(c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

The main advantage of reducing the process variation is that the chance of getting the defecting item will be reduced as we can see from the reduction which takes place from a to b from above.

7 0

4 years ago

Compare the ruler you made to an inch ruler.Describe how they are alike and how they are different?

Ainat [17]

Just compare the one you made to a normal ruler... Look at size, shape, etc. Simple.

4 0

3 years ago