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OleMash [197]
2 years ago
14

Which of these are statistical questions that you could ask to gather data on the types of transportation to school?

Mathematics
1 answer:
Pavlova-9 [17]2 years ago
7 0

Answer:

A statistical question is a question that can be answered by collecting data that vary. For example, “How old am I?” is not a statistical question, but “How old are the students in my school?” is a statistical question.

The closest ones are - B and E

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At a computer store, a customer is considering 9 different computers,10 different monitors,8 different printers and 2different s
Leto [7]

Answer:

The number of different computer systems possible is 1440.

Step-by-step explanation:

For each computer, there are 10 options of monitor.

For each monitor, there are 8 printers.

For each printer, there are 2 scanners.

There are 9 computers.

Determine the number of different computer systems possible.

9*10*8*2 = 1440

The number of different computer systems possible is 1440.

3 0
3 years ago
Which polynomial expression represents the perimeter of the rectangle? (3a + 4) (2a - 1)​
Katen [24]

Answer:

10a + 6

Step-by-step explanation:

3a + 4 + 2a - 1

5a + 3

2( 5a + 3)

10a + 6

7 0
3 years ago
For a moving object, the force acting on the object varies directly with the objects acceleration. When a force of 35 N acts on
Genrish500 [490]
<h3>Answer:  3 m/s^2</h3>

=======================================================

According to Newton's Second Law, we know that

F = m*a

where F is the force applied, m is the mass and 'a' is the acceleration.

We see that this is a direct variation equation for F and a, such that m is the constant of variation. It's similar to how y = kx is also a direct variation equation.

Plug in F = 35 and a = 5 to find m

F = ma

35 = m*5

35/5 = m

7 = m

m = 7

The object has a mass of 7 kg

Our equation F = ma updates to F = 7a

Now plug in the force F = 21 to find 'a'

F = 7a

21 = 7a

21/7 = a

3 = a

a = 3

The acceleration will be 3 m/s^2

Notice how a smaller force applied means that the acceleration has also gone down as well.

8 0
3 years ago
At lift-off, a rocket weighs 40.2 tons, including the weight of fuel. It is fired vertically, and the fuel is consumed at the ra
Thepotemich [5.8K]

Step-by-step explanation:

Below is an attachment containing the solution.

8 0
3 years ago
Uestion
Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}

now, let's check the other path, JM and KL

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}

so the red path will be  5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare

3 0
2 years ago
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