105.6 L/h = ___ L/min?

1 answer:

5 0

$(\frac{105.6\ L}{hour})(\frac{1\ hour}{60\ minutes})$

The hours cancel out and you divide 105.6 by 60 to get

1.76 L/min

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A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to fin

drek231 [11]

Answer:

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 28$

The alternate hypotesis is:

$H_{1} < 28$

The test statistic is:

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

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position.

This means that $n = 50, X = 26$

Assume the population standard deviation is 6.2 weeks.

This means that $\sigma = 6.2$

Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

We have to find the pvalue of Z, looking at the z-table, when $\mu = 28$ . It if is lower than 0.05, it provides evidence.

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

$z = \frac{26 - 28}{\frac{6.2}{\sqrt{50}}}$

$z = -2.28$

$z = -2.28$ has a pvalue of 0.0113 < 0.05.

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

5 0

2 years ago

The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of

valina [46]

Answer:

(221.39, 300.61) and (255.2223, 266.7777)

Step-by-step explanation:

Given that X, the lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days

Middle 98% would lie on either side of the mean with probability ±2.33 in the std normal distribution on either side of 0

Corresponding we have x scores as

Between $261-2.33*17$ and $261+2.33*17$