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Harman [31]
3 years ago
7

a house has a rectangular tile kitchen that is 7 tiles wide and 8 tiles long. It also has a rectangular dining room that is 9 ti

les wide and 10 tiles long. How many tiles are used for the floors of the kitchen and dining room?
Mathematics
1 answer:
Eduardwww [97]3 years ago
7 0
7*8 = 56
9*10 = 90 add them
ANSWER = 146 tiles
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A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.
AVprozaik [17]

Answer:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean  is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

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The standard deviation of pulse rates of adult males is more than 11 bpm. For the random sample of 153 adult males the pulse rat
Alenkasestr [34]

Answer: 160.4040

Step-by-step explanation:

Here , the claim is "The standard deviation of pulse rates of adult males is more than 11 bpm." , i.e. \sigma>11

We use Chi-square test statistic for the test statistic to test the standard deviations :

\chi^2=\dfrac{(n-1)s^2}{\sigma^2} , where s= sample standard deviation , \sigma = population standard deviations and n = sample size.

As per given , n=153 , s= 11.3

Then, Required test statistic will be :

\chi^2=\dfrac{(153-1)(11.3)^2}{11^2}=\dfrac{152\times127.69}{121}\approx160.4040

Hence, the value of the test statistic. =  160.4040

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Arte-miy333 [17]

Answer:

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Step-by-step explanation:

the slope of the first line is 2

to get the line perpendicular, you swap the numerator and denominator, then if positive make negative and the other way around

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Answer:

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Step-by-step explanation:

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