If L is parallel to M find the value of x and the value of y.

9 divided by 204 show your work

andrew-mc [135]

Dude, you could just use a calculator but the answer is 51.

6 0

3 years ago

Read 2 more answers

you are asked to draw a triangle with side length of 10in,7in and 2in how many triangles like this can you draw

Aleks [24]

None because two sides of the triangle added together have to be greater than the other side

5 0

3 years ago

sons in the NHL,nts than anyone whoccomplished thiser games thanHer. Here are theRg each season:74, 45, 81, 48, 80,Working for a

vesna_86 [32]

Step 1:

In descriptive statistics, a box plot or boxplot (also known as box and whisker plot) is a type of chart often used in explanatory data analysis. Box plots visually show the distribution of numerical data and skewness through displaying the data quartiles (or percentiles) and averages.

Step 2:

To sketch a boxplot, you will need to determine the following:

Minimum

Lower quartile

Median

Upper quartile

Maximum

Step 3:

USA

First arrange the data from the least to the greatest.

14 , 26, 33, 34, 35, 38, 40, 41, 44, 46, 46, 49, 53, 66, 70.

Minimum = 14

$\begin{gathered} \text{Lower quartile position = }\frac{1}{4}(n+1)^{th} \\ \text{= }\frac{15+1}{4}\text{ = }\frac{16}{4}=4^{th} \\ Q_1\text{ = lower quartile = 34} \end{gathered}$

Median = 41

$\begin{gathered} \text{Upper quartile position = }\frac{3}{4}(n+1)^{th} \\ =\text{ }\frac{3\text{ }\times(15+1)}{4} \\ =12^{th} \\ Q_3\text{ = upper quartile = 49} \end{gathered}$

Maximum = 70

Interquartile range IQR = 49 - 34 = 15

Canada

22, 28, 31, 31, 35, 35, 36, 39, 40, 40, 42, 49, 53, 63, 68

Minimum = 22

$\begin{gathered} Q1\text{ Position = }\frac{(n+1)^{th}\text{ }}{4}\text{ = }\frac{16}{4}th=4^{th} \\ Q_1\text{ = 31} \end{gathered}$

Median = 39

$\begin{gathered} UpperquartileQ_{_3\text{ }}\text{ = 49} \\ \end{gathered}$

Maximum = 68

Interquartile range = 49 - 31 = 18

8 0

1 year ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

Deffense [45]

Answer:

a) $17-2.26\frac{1.9}{\sqrt{10}}=15.64$

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$

So on this case the 95% confidence interval would be given by (15.64;18.36)

b) 1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Step-by-step explanation:

Part a: What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places)

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".

Solution to the problem

We have the following data:

$\bar x= 17$ represent the sample mean

$s = 1.9$ represent the sample deviation

n =10 represent the sample size

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

$df=n-1=10-1=9$

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 table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.26$

Now we have everything in order to replace into formula (1):

$17-2.26\frac{1.9}{\sqrt{10}}=15.64$

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$

So on this case the 95% confidence interval would be given by (15.64;18.36)

Part b

1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

5 0

3 years ago

What are the coordinates of a vertex of the parabola with the equation y=x2-3

neonofarm [45]

Answer:

Step-by-step explanation:

Vertex is (0,-3

6 0

3 years ago