What is the (GCF) of the sum 44+40=??

Name the sampling method used in each of the following situations:

elixir [45]

Answer:

a) Convenient

b) Cluster

c) Stratified

d) Systematic

e) Random

Step-by-step explanation:

Samples may be classified as:

Convenient: Sample drawn from a conveniently available pool.

Random: Basically, put all the options into a hat and drawn some of them.

Systematic: Every kth element is taken. For example, you want to survey something on the street, you interview every 5th person, for example.

Cluster: Divides population into groups, called clusters, and each element in the cluster is surveyed.

Stratified: Also divides the population into groups. However, then only some elements of the group are surveyed.

Situation a:

Asks all from a single group(sitting passengers), that is, convenint sample.

Situation b:

First she separates the students by row, then from each row, she chooses all students. So this is Cluster sampling.

Situation c:

The manager separates the stores by location, and then in each location, random customers are chosen. This is an example of Stratified sampling.

Situation d:

Marks for every fourth patron, that is, every 4th element. So Systematic sampling.

Situation e:

People chosen at random, so Random sampling.

6 0

3 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

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

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

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago

Subtract.

3241004551 [841]

Answer:

405

Step-by-step explanation:

5 0

2 years ago

Two lines that have the same slope do not intersect at any points

Crazy boy [7]

if you are looking for the solution then is <em>Infinitely many solutions</em>

4 0

2 years ago

Does anybody know how to do time with exponential decay

My name is Ann [436]

<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

$b_n=0.12\times b_{n-1}$

Why does this work? Initially, you start with 500 mL of air that you breathe in, so $b_1=500\text{ mL}$ . After the second breath, you have 12% of the original air left in your lungs, or $b_2=0.12\timesb_1=0.12\times500=60\text{ mL}$ . After the third breath, you have $b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}$ , and so on.

You can find the amount of original air left in your lungs after $n$ breaths by solving for $b_n$ explicitly. This isn't too hard:

$b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots$

and so on. The pattern is such that you arrive at

$b_n=0.12^{n-1}b_1$

and so the amount of air remaining after $50$ breaths is

$b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}$

which is a very small number close to zero.</span>

5 0

3 years ago