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2 years ago

The equations are given below.

Mathematics

1 answer:

amid [387]2 years ago

5 0

Answer:

neither

Step-by-step explanation:

they are parallel when they have the same slope and perpendicular when the slope is completely different

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Consider a sampling distribution with p equals 0.15p=0.15 and samples of size n each. Using the appropriate formulas, find the

Grace [21]

Answer:

$a.\ \mu_p=750\ \ , \sigma_p=0.005\\\\b.\ \mu_p=150\ \ , \sigma_p=0.0113\\\\c.\ \mu_p=75\ \ , \sigma_p=0.0160$

Step-by-step explanation:

a. Given p=0.15.

-The mean of a sampling proportion of n=5000 is calculated as:

$\mu_p=np\\\\=0.15\times 5000\\\\=750$

-The standard deviation is calculated using the formula:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\=\sqrt{\frac{0.15(1-0.15)}{5000}}\\\\=0.0050$

Hence, the sample mean is μ=750 and standard deviation is σ=0.0050

b. Given that p=0.15 and n=1000

#The mean of a sampling proportion of n=1000 is calculated as:

$\mu_p=np\\\\=1000\times 0.15\\\\\\=150$

#-The standard deviation is calculated as follows:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{1000}}\\\\\\=0.0113$

Hence, the sample mean is μ=150 and standard deviation is σ=0.0113

c. For p=0.15 and n=500

#The mean is calculated as follows:

$\mu_p=np\\\\\\=0.15\times 500\\\\=75$

#The standard deviation of the sample proportion is calculated as:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{500}}\\\\\\=0.0160$

Hence, the sample mean is μ=75 and standard deviation is σ=0.0160

4 0

3 years ago

A taste test asks people from Texas and California which pasta they prefer,

melamori03 [73]

Answer:

Yes, they are independent because P(Texas) ≈ 0.45 and P(Texas/brand A) ≈ 0.45.

Step-by-step explanation:

We are given that a taste test asks people from Texas and California which pasta they prefer, brand A or brand B. The table is given in the question.

A person is randomly selected from those tested.

And we have find that are being from Texas and preferring brand A independent events or not.

Firstly, we know that these two events will be independent when;

P(Texas) = P(Texas/brand A)

Now, P(Texas) = $\frac{\text{Number of people from Texas}}{\text{Total number of people}}$

= $\frac{125}{275}$ ≈ 0.45

Also, P(Texas/brand A) = $\frac{P(Texas \bigcap Brand A)}{P(Brand A)}$

= $\frac{\text{Number of people from Texas and preferring brand A}}{\text{Number of people preferring brand A}}$

= $\frac{80}{176}$ ≈ 0.45

Therefore, being from Texas and preferring brand A are independent events because P(Texas) ≈ 0.45 and P(Texas/brand A) ≈ 0.45.

6 0

2 years ago

Can anyone help with this assignment please

Lera25 [3.4K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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 $\bar X$

8 0

3 years ago

Find the GCF of the following terms. <br> 16xy^3 , 20x^2y^2 , -4x^2y

Ivanshal [37]

$16xy^3=4xy\cdot4y^2\\\\20x^2y^2=4xy\cdot5xy\\\\-4x^2y=4xy\cdot(-x)\\\\GCF(16xy^3;\ 20x^2y^2;\ -4x^2y)=4xy$

7 0

3 years ago