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4 years ago

How many people were included in the sample shown below?

Mathematics

2 answers:

Rama09 [41]4 years ago

8 0

Answer:

whats the answer

Step-by-step explanation:

Svetlanka [38]4 years ago

5 0

Answer:

7 people

Step-by-step explanation:

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Need help ( 8th grade math)

kifflom [539]

It would be -3/1. This is because you find where it crosses the Y axis which is at 4. Then you go down three and over one to give you the next even point. Which going down to the right means it’s negative. Hope this helps! Please mark branlist if possible!!

6 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%7B4m%7D%5E%7B2%7D%20%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%7Bm%7D%5E%7B3%7D%20%20%5C

olchik [2.2K]

Step-by-step explanation:

I've shown the process in the picture.

hope u understand

8 0

3 years ago

A square is removed from the right hand corner

denis-greek [22]

Answer:

C im pretty sure

Step-by-step explanation:

Because the square taken out has a 4x4 area which is 16 and the area of the whole square used to be 14x14 which is 196 and if u subtract 16 from that you get 180

5 0

3 years ago

Read 2 more answers

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....