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4 years ago

Round 132,654 to the nearest one thousand

2 answers:

8 0

If you are rounding to the nearest thousands place, you look at the hundreds place. The hundreds place is a 6. Since 6 is larger than 4, you would round up. So the answer is 133,000.

grin007 [14]4 years ago

6 0

Well the 2 is in the thousands and the number before it is six and because of the number before 2 happens to be over 5 the 2 changes into a 3 thus making your answer 133,000

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Five observations taken for two variables follow.

I am Lyosha [343]

Answer:

a) Figure attached

b) If we see the scatter plot we can conclude that the possible relation between x and y is linear and with a positive correlation since when the values of x increases the values for y increases as well.

c) $Cov (X,Y) = \frac{\sum_{i=1}^n (x_i -\bar X)(y_i -\bar Y)}{n-1}$

We can find the numerator like this:

$\sum_{i=1}^5 (6-16)(6-10)+(11-16)(9-10)+(15-16)(6-10)+(21-16)(17-10)+(27-16)(12-10)=106$

And then:

$Cov(X,Y) = \frac{106}{5-1}=26.5$

d) $r=\frac{5(906)-(80)(50)}{\sqrt{[5(1552) -(80)^2][5(586) -(50)^2]}}=0.693$

Step-by-step explanation:

Part a

For this part we use excel in order to create the scatterplot and we got the result on the figure attached.

Part b

If we see the scatter plot we can conclude that the possible relation between x and y is linear and with a positive correlation since when the values of x increases the values for y increases as well.

Part c

The sample covariance is defined as:

$Cov (X,Y) = \frac{\sum_{i=1}^n (x_i -\bar X)(y_i -\bar Y)}{n-1}$

We can find the numerator like this:

$\sum_{i=1}^5 (6-16)(6-10)+(11-16)(9-10)+(15-16)(6-10)+(21-16)(17-10)+(27-16)(12-10)=106$

And then:

$Cov(X,Y) = \frac{106}{5-1}=26.5$

Part d

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=5 $\sum x = 80, \sum y = 50, \sum xy = 906, \sum x^2 =1552, \sum y^2 =586$

$r=\frac{5(906)-(80)(50)}{\sqrt{[5(1552) -(80)^2][5(586) -(50)^2]}}=0.693$

So then the correlation coefficient would be r =0.693

3 0

3 years ago

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medici

Ilia_Sergeevich [38]

Answer:

$4600-1.646\frac{800}{\sqrt{1800}}=4568.96$

$4600+1.646\frac{800}{\sqrt{1800}}=4631.04$

We are confident at 90% that the true mean for the amount spent on medicines is between (4568.96 and 4631.04)

Step-by-step explanation:

Information given

$\bar X=4600$ represent the sample mean for the amount spent on medicines

$\mu$ population mean

s=800 represent the sample standard deviation

n=1800 represent the sample size

Solution

The confidence interval for the true population mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

We can calculate the degrees of freedom with:

$df=n-1=1800-1=1799$

We know that the Confidence level is 0.90 or 90%, the value of significance is $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,1799)".And we see that $t_{\alpha/2}=1.646$