Find the slope of the line through the given points.<br> (-2,7) and (2,-5)

The area of the rug is 4 15/36<br> The rug is 2 3/4 meters in height <br> How wide is the rug

sineoko [7]

I assume the rug is rectangle.

The area of rectangle is width (W) multiply by hight (H)

$A=4\frac{15}{36}=4\frac{5\cdot3}{12\cdot3}=4\frac5{12}\\\\H=2\frac34\\\\\\A=W\cdot H\\\\4\frac{5}{12}=W\cdot2\frac34\\\\4\frac5{12}:2\frac34=W\\\\ \frac{4\cdot12+5}{12}:\frac{2\cdot4+3}{4}=W\\\\ \frac{53}{12}:\frac{11}4=W\\\\ \frac{53}{12}\cdot\frac4{11}=W\\\\ \frac{53\cdot4}{3\cdot4\cdot11}=W\\\\W=\frac{53}{33}\\\\W=1\frac{20}{33}$

The rug is $1\frac{20}{33}$ meters wide

4 0

3 years ago

Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student

AysviL [449]

Answer:

The mean calculated for this case is $\bar X=584$

And the 95% confidence interval is given by:

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=584$

The sample deviation calculated $s=86.776$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$