I really need your help please

Which expression is equivalent to (2^1/2 2^3/4)^2

dimaraw [331]

Answer:

$\sqrt{2^5}$

Step-by-step explanation:

$2^{1/2}$ × $2^{3/4}$ = $2^{5/4}$

( $2^{5/4}$ )² = $2^{5/2}$ = $\sqrt{2^5}$

6 0

3 years ago

Read 2 more answers

The mean rent of a 3-bedroom apartment in Orlando is $1300. You randomly select 10 apartments around town. The rents are normall

lana [24]

Answer:

96.49% probability that the mean rent is more than $1100

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 1300, \sigma = 350, n = 10, s = \frac{350}{\sqrt{10}} = 110.68$

What is the probability that the mean rent is more than $1100?

This is 1 subtracted by the pvalue of Z when X = 1100. So

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

By the Central Limit Theorem

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

$Z = \frac{1100 - 1300}{110.68}$

$Z = -1.81$

$Z = -1.81$ has a pvalue of 0.0351

1 - 0.0351 = 0.9649

96.49% probability that the mean rent is more than $1100

7 0

4 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$