Systematic random sampling is used to interveiw residents in 25% of 80 apartments in a building. The sampling interval would be

What is another way to write 27/5

kow [346]

Answer 5.4

Step-by-step explanation:

8 0

3 years ago

Triangle XYZ is transformed to create triangle X'Y'Z'. The side lengths of both triangles are shown.

tino4ka555 [31]

Answer:

Yes, the pre-image and image have the same side length measures

Step-by-step explanation:

we know that

A rigid transformations not change the size or shape of the objects.

Reflections, translations, rotations, and combinations of these three transformations are "rigid transformations.

In this problem

The pre-image and image have the same side length measures

therefore

Is a rigid transformation

8 0

3 years ago

Read 2 more answers

∆ABC is translated 6 units up and 3 units left to create ∆A'B'C'. If vertex A is at (-1, 2) and vertex B is at (1, 5), then vert

Paladinen [302]

So we want to know the new coordinates for the vertex A'(x,y) if we know that the vertex A is at A(-1,2) and vertex B is at B(1,5) and that the triangle ABC is translated 6 units up and 3 units left. So the method is simply to add units 6 to x and 3 to y of A to get A'. Going left means we need to go to negative x direction and going up means we need to go to positive y direction. So: A'(-1-3,2+6) and that is: A'(-4,8).

4 0

3 years ago

Read 2 more answers

Three Times what Equals 75

allochka39001 [22]

Let's define what as 'x'

3x = 75

Divide both sides by 3.

3x/3 = 75/3

x = 75/3

x = 25

Three times twenty five equals 75.

Your final answer is twenty five or 25.

3 0

3 years ago

Read 2 more answers

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

$\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}$

$\frak {\huge{ \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

$\sf let \: f(x) = { \tan }^{ - 1} x \\ \\ \sf f(x + h) = { \tan}^{ - 1} (x + h)$

$\sf f'(x) = \frac{f(x+h) - f(x) }{h}$

$\sf \implies \lim_{ h \to 0 } \frac{ { \tan }^{ - 1}(x + h) - { \tan}^{ - 1}x }{h} \\ \\ \\ \sf \implies \lim_ {h \to 0} \frac{ { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}$

By using

$\sf { \tan}^{ - 1} x - { \tan}^{ - 1} y = \\ \sf { \tan}^{ - 1} \frac{x - y}{1 + xy} formula$

$\sf \implies \large \lim_{h \to0 } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{h} \\ \\ \\ \sf \implies \large{\lim_{h \to0} } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } \times (1 + hx + {x}^{2} )} \\ \\ \\ \sf \implies \large \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } } + \lim_{h \to0} \frac{1}{1 + hx + {x}^{2} }$

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

6 0

2 years ago