Can someone help me

Find the gradient of the line segment between the points (-5,2) and (4,3) give your answer in the simplest form

Leokris [45]

Answer: The gradient of the line segment between the points (-5,2) and (4,3) is $\dfrac{1}{9}$ .

Step-by-step explanation:

Formula for gradient of a line segment :

$\text{Gradient}=\dfrac{\text{Difference between y-coordinates}}{\text{Difference between x-coordinates}}$

Given points of the line segment: (-5,2) and (4,3)

$\text{Gradient} =\dfrac{3-2}{4-(-5)}$

$\text{Gradient} =\dfrac{1}{4+5)}$

$\text{Gradient} =\dfrac{1}{9}$

Therefore , the gradient of the line segment between the points (-5,2) and (4,3) is $\dfrac{1}{9}$ .

7 0

3 years ago

Please someone help answer correctly !!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!

kogti [31]

Answer:

z = 44

Step-by-step explanation:

180 - (44+52) = 180 - 96 = 84º

180 - 84 = 96º

180 - (96 + 40) = z

180 - 136 = z

z = 44

5 0

3 years ago

Shards of clay vessels were put together to reconstruct rim diameters of the original ceramic vessels found at the Wind Mountain

Otrada [13]

Answer:

a) 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)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)^2}{n-1}}$ (3)

The mean calculated for this case is

The sample deviation calculated $s=3.46 \approx 3.5$

b) $15.8-1.383\frac{3.5}{\sqrt{10}}=14.27$

$15.8+1.383\frac{3.5}{\sqrt{10}}=17.33$

So on this case the 80% confidence interval would be given by (14.27;17.33)

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

Part a

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)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)^2}{n-1}}$ (3)

The mean calculated for this case is

The sample deviation calculated $s=3.46 \approx 3.5$

Part b

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 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,9)".And we see that $t_{\alpha/2}=1.383$