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LekaFEV [45]
3 years ago
12

The accompanying observations are numbers of defects in 25 1-square-yard specimens of woven fabric of a certain type: 4, 7, 5, 2

, 3, 1, 9, 3, 4, 3, 5, 7, 3, 2, 2, 4, 7, 3, 2, 5, 5, 1, 4, 4, 5. Construct a c chart for the number of defects. (Round your answers to two decimal places.)
x double bar =
LCL =
UCL =
Mathematics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

x double bar = 4

LCL = - 2

UCL = 10

Step-by-step explanation:

The C - chart :

The control limit = xbar

xbar = mean = ΣX / n

The upper and lower control limit :

Xbar ± 3√xbar

Xbar = Σx / n = 100 / 25 = 4

Hence,

x double bar = 4

LCL = Xbar - 3√xbar = 4 - 3√4 = 4 - (3*2)

LCL = 4 - 6 = -2

UCL = Xbar + 3√xbar = 4 + 3√4 = 4 + (3*2)

LCL = 4 + 6 = 10

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Solve the compound inequality, then correctly fill in the blanks below.<br> -3 2-5 &lt;2<br> I
lesantik [10]

Answer:

hold my picolr

Step-by-step explanation:

7 0
3 years ago
Hao bought a 16-pound bag of dog food. If his dog eats pound of food every day, for how many days will the bag of dog food last?
densk [106]

Answer:

uh 1 pound per day 16 pounds so 16 days, i guess

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
It takes a printer 2 1/2 min to print 24 black-and-white pages and 4 color pages. It takes the same printer 3 1/4 min to print 3
Paul [167]
Multiply the first equation by 1.5, to get:
(1.5)(24b + 4c) = 1.5(2.5)
36b + 6c = 3.75
Subtract the second equation from this:
6b = 1/2
b = 1/12
Substituting into the first equation:
24b + 4c = 2.5
24*(1/12) + 4c = 2.5
2 + 4c = 2.5
4c = 0.5
c = 1/8
Therefore, it takes 1/12 minute to print a black-and-white page, and 1/8 minute to print a colored page.
4 0
3 years ago
Read 2 more answers
A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv
lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

            of  novels that seniors read during their time in college. An exit

            survey was conducted with a random sample of 9 seniors. The

            sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

              been met, which of the following is a 94.645% confidence

              interval for the population mean number of novels read by

              all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

                                = n - 1

                                = 9 - 1

                                = 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

   =1 - 0.94645

  =0.05355 ≈ 0.05

⇒α = 0.5

Now,

\frac{\alpha}{2} = \frac{0.05}{2}

  = 0.025

Then,

t_{\frac{\alpha}{2}, df }  = 2.306

∴ we get

Population mean = x⁻ ± t_{\frac{\alpha}{2}, df } ×\frac{s}{\sqrt{n} }

                           = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

⇒Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

3 0
3 years ago
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MakcuM [25]
This is a typical work problem in algebra. The approach to this is using the equation: Rt = 1, where R is the rate per person working, t is the amount of time worked by an individual. All their rates must equal to 1. The solution is as follows:

12/x + 15/x = 1
Solving for x,
x = 27 hours

So, if they work together, they can finish the work in 27 hours.
6 0
3 years ago
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