Number of pennies, nickels, dimes, quarters, and other types of currency contained in a piggy bank:

Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a

omeli [17]

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

$\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287$

This probability is 1 subtracted by the pvalue of Z when X = 900.

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

By the Central Limit Theorem

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

$Z = \frac{900 - 880}{9.1287}$

$Z = 2.19$

$Z = 2.19$ has a pvalue of 0.9857

1 - 0.9857 = 0.0143

1.43% probability that the sample mean rent price will be greater than $900

8 0

3 years ago

Tom determines the system of equations below has two solutions, one of which is located at the vertex of the parabola.

Rus_ich [418]

Answer:

b must equal 7 and a second solution to the system must be located at (2, 5).

Step-by-step explanation:

Rearranging the first equation:

y = (x - 3)^2 + 4

From this we see that the vertex is at the point (3,4).

So one solution of equation 2 is (3 ,4).

Substituting in equation 2:

4 = -3 + b

b = 7.

So equation 2 is y = - x + 7.

Now we check if (2, 5) is on this line:

5 = -2 + 7 = 5 , therefore (2, 5) is on this line.

Verifying if (2, 5) is also on y = (x - 3)^2 + 4:

5 = (2 - 3)^2 + 4 = 1 + 4 = 5

- so it is. and a second solution to the system is (2, 5).

5 0

3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

Can someone pls help, I know this is kind of a lot but I’ve been stuck on this for a while now.

Greeley [361]

9514 1404 393

Answer:

(a, b, c) = (-0.425595, 11.7321, 2.16667)

f(x) = -0.425595x² +11.7321x +2.16667

f(1) ≈ 13.5

Step-by-step explanation:

A suitable tool makes short work of this. Most spreadsheets and graphing calculators will do quadratic regression. All you have to do is enter the data and make use of the appropriate built-in functions.

Desmos will do least-squares fitting of almost any function you want to use as a model. It tells you ...

a = -0.425595

b = 11.7321

c = 2.16667

so

f(x) = -0.425595x² +11.7321x +2.16667

and f(1) ≈ 13.5

_____

<em>Additional comment</em>

Note that a quadratic function doesn't model the data very well if you're trying to extrapolate to times outside the original domain.

3 0

3 years ago

What transformation was made for this triangle?

Musya8 [376]

Answer:

Dilation

Step-by-step explanation:

Dilation is when a shape is made bigger but the form of the shape does not change. So its enlarged.

7 0

3 years ago