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3 years ago

7

Ten samples of a process measuring the number of returns per 100 receipts were taken for a local retail store. The number of ret

urns were 10, 9, 11, 7, 3, 12, 8, 4, 6, and 11. Find the standard deviation of the sampling distribution for the p-bar chart. 0.0863 0.081 0.0273 There is not enough information to answer the question. 8.1

1 answer:

Tatiana [17]3 years ago

3 0

Answer:

0.0273

Step-by-step explanation:

np n

10 100

9 100

11 100

7 100

3 100

12 100

8 100

4 100

6 100

11 100

pbar=sumnp/sumn

pbar=10+9+11+7+3+12+8+4+6+11/10+10+10+10+10+10+10+10+10+10

pbar=81/1000

pbar=0.081

nbar=sumn/k=1000/10=100

$Standard deviation for pbar chart=\sqrt{\frac{pbar(1-pbar)}{nbar} }$

$Standard deviation for pbar chart=\sqrt{\frac{0.081(0.919)}{100} }$

$Standard deviation for pbar chart=\sqrt{\frac{0.0744}{100} }$

$Standard deviation for pbar chart=\sqrt{0.0007444 }$

Standard deviation for p-chart=0.0273

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What vaule of x makes th question true? 3x+2(x-5)=50 a. 8 b.9 c.11 d.12

Ainat [17]

Hey Tierra7l,

Lets solve your equation together step by step.

$3x+2\left(x-5\right)=50$

$\mathrm{Expand}\:2\left(x-5\right):\quad 2x-10$

$\mathrm{Distribute\:parentheses\:using}:\quad \:a\left(b+c\right)=ab+ac$

$a=2,\:b=x,\:c=-5$

$2\cdot \:x+2\left(-5\right)$

$\mathrm{Apply\:minus-plus\:rules}$

$\:\:+\left(-a\right)=-a$

$2x-2\cdot \:5$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:5=10$

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$\mathrm{Add\:similar\:elements:}\:3x+2x=5x$

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$\mathrm{Add\:}10\mathrm{\:to\:both\:sides}$

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AnthrαX <span> </span>

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3 years ago

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finlep [7]

Answer:

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Step-by-step explanation:

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2 years ago

Fill in the missing information. Tim Worker is doing his budget. He discovers that the average miscellaneous expense is $45.00 w

nlexa [21]

Answer:

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And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

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And replacing we got:

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We can find the probability of interest using the normal standard table and with the following difference:

$P(-0.4$

Step-by-step explanation:

Let X the random variable who represent the expense and we assume the following parameters:

$\mu = 45, \sigma 16$

And for this case we want to find the percent of his expense between 38.6 and 57.8 so we want this probability:

$P(38.6$

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

We can find the probability of interest using the normal standard table and with the following difference:

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nadya68 [22]

Answer:

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