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3 years ago

6

A machine packs boxes at a constant rate of 2/3 of a box every 1/2 minute. What is the number of boxes per minute that the machi

ne packs?

1 answer:

Tju [1.3M]3 years ago

8 0

Answer:

Basicallly, if the machine packs 2/3 of a box every 1/2 min, then the machine will pack twice as many boxes for 1 min, since 1/2 multiplied with 2 is 1.

2/3 multiplied with 2 is 4/3.

Therefore, the machine can pack 4/3 boxes per minute.

(what my research says)

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3 years ago

Read 2 more answers

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

a

$P(\= X \ge 51 ) =0.0062$

b

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

So

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

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2 years ago