You move down 6 units. You end at (-5, -5). Where did you start?

It takes 1 gallon of paint to cover 18 fence boards. Assuming you can only buy whole gallons of paint at the store, how many gal

ankoles [38]

Answer:

9 gal

Step-by-step explanation:

150/18

8.333333333333334

if we round that to the greater whole gallon we get 9 gallons

5 0

3 years ago

Using traditional methods it takes 92 hours to receive an advanced flying license. A new training technique using Computer Aided

DanielleElmas [232]

Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

$H_0: \mu = 92$

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

$H_1: \mu > 92$

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.

92 is tested at the null hypothesis:

This means that $\mu = 92$

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that $n = 70, X = 93, \sigma = \sqrt{36} = 6$

Value of the test statistic:

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

$z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}$

$z = 1.39$

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

8 0

3 years ago

You just started working for a company, where should you first look to find a confidentiality policy concerning the information

olasank [31]

It would be D because ur an employee and u need a handbook to see what ur handelling

4 0

3 years ago

The mean yearly rainfall in Sydney, Australia, is about 134 mm and the standard deviation is about 66 mm ("Annual maximums of,"

svetlana [45]

Answer:

At least 202.44 mm in the top 15%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 134, \sigma = 66$

How many yearly mm of rainfall would there be in the top 15%?

At least X mm.

X is the 100-15 = 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037.

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

$1.037 = \frac{X - 134}{66}$

$X - 134 = 66*1.037$

$X = 202.44$

At least 202.44 mm in the top 15%.

5 0

3 years ago

A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip

olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

$\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t + \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V$

Simplifying gives;

$\dfrac{(2\cdot t +7) \cdot V}{16} = V$

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0

3 years ago

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