Find PQ if possible

You look at real estate ads for houses in Naples, Florida. There are many houses ranging from $200,000 to $500,00 in price. The

cricket20 [7]

Answer:

c) skewed to the right.

Step-by-step explanation:

We need to remember that is a distribution is skewed to the right then we have the following condition satisfied:

$Mode< Median$

And if is skewed to the left then we have:

$Mean$

If the distribution is symmetric we need to satisfy:

$Mean = Median=Mode$

For this case since we have most of the values between 200000 and 500000 when we put atypical values like 15000000 that would affect the sample mean and on this case the sample mean would larger than the sample median because the median is a robust measure of central tendency not affected by outliers.

So for this special case we can say that the $Mean>Median$ . And probably the median would be higher than the mode so then we can conclude that the best answer for this case would be:

c) skewed to the right.

5 0

3 years ago

Find m. Please guys I really need your help :(

vredina [299]

Answer:

$m=\sqrt{2}$ inches

Step-by-step explanation:

From the diagram, it is given that the triangle is a right angle isosceles triangle (angles are 45°, 45° and 90°). This means that the two shorter sides of the triangle are equal in length and that the side $m$ can be found by using the Pythagorean theorem $a^{2}+b^{2}=c^{2}$ .

$m^{2}+m^{2}=2^{2}$ (The two shorter sides of the triangle are equivalent)

$2m^{2}=4$

$m^{2}=2$

$m=\sqrt{2}$ inches

I hope this helps :)

8 0

3 years ago

Same receive the weekly paycheck here's a copy of one of his pay statements if there are four pages per month what is Sam's mont

Lady_Fox [76]

$654.48 i believe. maybe

4 0

3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

Write an equation for g(x), a translation of f(x) 4 units to the right and 3 units down.

Hoochie [10]

I’m pretty sure it’s c) I’m sorry if it’s wrong :)

3 0

3 years ago