Write an equation in slope-intercept form for the line with slope 7 and y-intercept -5

Find the distance between the points with the given coordinates (-4, 9) and (1, -3).

Dmitriy789 [7]

Answer: 13

Step-by-step explanation:

The formula for finding distance between two points is given by :

D = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$x_{1}$ = -4

$x_{2}$ = 1

$y_{1}$ = 9

$y_{2}$ = -3

Substituting the values into the formula , we have :

D = $\sqrt{(1 -(-4))^{2} +(-3-9)^{2}}$

D = $\sqrt{(5)^{2}+(-12)^{2}}$

D = $\sqrt{25+144}$

D = $\sqrt{169}$

D = 13

Therefore : the distance between the points is 13

7 0

3 years ago

One indicator of an outlier is that an observation is more than 2.5 standard deviations from the mean. Consider the data value 8

sergij07 [2.7K]

Answer:

a) No, since 80 is less than 2.5 standard deviations above the mean; b) Yes, since 80 is more than 2.5 standard deviations above the mean.

Step-by-step explanation:

For part a,

The mean is 70 and the standard deviation is 5. This means that 2 standard deviations above the mean is 70+5+5 = 80.

This means that 80 is not an outlier, as it is not more than 2.5 standard deviations above the mean.

For part b,

The mean is 70 and the standard deviation is 3. This means that 2 standard deviations above the mean is 70+3+3 = 76; 3 standard deviations above the mean is 70+3+3+3 = 79. 80 is more than this, so yes, it is an outlier.

4 0

4 years ago

Find two numbers that differ by 8 and whose square roots differ by 1. Those two numbers are ___ and ____

Anna11 [10]

Answer: 12.25 and 20.25

Step-by-step explanation:

20.25 - 12.25 = 8

Square root of 20.25 is 4.5

Square root of 12.25 is 3.5

4.5 - 3.5 = 1

Hope this helped

7 0

4 years ago

Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y =

Black_prince [1.1K]

Answer:

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

Step-by-step explanation:

First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line $x + 2y= 0, x\geq 0$

If we choose $x = 2$ (we can do so because of the condition $x \geq 0$ , which means that any positive value is suitable for $x$ ), then we have

$2 +2y = 0\implies 2 = -2y \implies y = -1$

Therefore, the terminal side of the angle $\theta$ is passing through the origin and the point $(2,-1)$ and now we can draw it.

The angle $\theta$ is presented below.

The distance of the point $(2,-1)$ from the origin equals

$r = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

Now, we can determine the values of the six trigonometric function, by using their definitions.

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

6 0

4 years ago

Use the graph to answer parts a and b below.

slega [8]

The correct answer is (4, 90)

4 0

3 years ago

Read 2 more answers