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german
3 years ago
8

are there earned $136 in three weeks. He goes back to school and one week needs at least $189 to buy new coat. How much more mon

ey does he need?
Mathematics
1 answer:
const2013 [10]3 years ago
5 0
He needs exactly $53
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A. 200.625 minutes   B. 184.166 minutes  C. 200 minutes  (round these answers like the question asked) 4. Plan A is the best because you get the most minutes for your money. I will explain my work for A, but if you need to explain the rest just ask. So, if you have to pay $3.95 every month no matter what and $0.08 for every minute you talk you can write the equation as, Cost = 3.95 + 0.08m (m stands for minutes), and the cost is 20$ then the equation is 20 = 3.95 + 0.08m, subtract 3.95 from the right side to make it 16.05 = 0.08m, then divide everything by 0.08 to get m, which gives you m= 200.625 
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3 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:

a)

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.

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4 years ago
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Last year Henry made 32 one cups serving of soup school party two times
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64 one cup servings your question seems incomplete

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Answer:

y=4

Step-by-step explanation:

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divide both sides by 4

4=y

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4 years ago
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