All categories

3 years ago

Pls help will give brainliest

Mathematics

1 answer:

weeeeeb [17]3 years ago

6 0

Answer:

<h2>Both classes will have the same value for the third quartile </h2>

Step-by-step explanation:

This is it because the 3rd quartile is where the maximum or whisker is. Hope this helps!

You might be interested in

The table below shows the ages of 50 people on a bus.

emmasim [6.3K]

Step-by-step explanation:

The table is missing pls attach table so that I can answer your question thank you so much for cooperating with me

7 0

2 years ago

A board measures 34.1 cm. What is the greatest possible error allowed for the measurement?

gladu [14]

Greatest error is half smallest measuring unit. It's measured to 0.1 cm , half that is 0.05 cm answer D.

7 0

3 years ago

How to find the surface area of the open net

Kruka [31]

To find the surface area of the open net, you’ll have to find the area of the faces
So, since there are 6 boxes you’ll have to find the area of the faces and add it up to find the surface area
10x4=40x4=160
3x4=12x2=24
160+24=184

Hope this helps ;)

5 0

3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

3 years ago

Help me!!!!!! Hanger math..

Gekata [30.6K]

Answer:x=3

Step-by-step explanation:

Equation:

5x+2=17

5x=15

x=3

5 0

3 years ago

Read 2 more answers