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Furkat [3]
3 years ago
12

David wants to make a box plot showing histeam's points for the year. The median score was 7, first quartile was 4, and third qu

artile was 10. The minimum was 2 and the maximum was 20. Explain how David can draw the box plot.

Mathematics
1 answer:
vichka [17]3 years ago
8 0
A dot above 7, 4, 10, 2, and 20. Draw a box around the median, first quartile, and and the third quartile. Then connect the box to the minimum and maximum by drawing a line. Should look like this:

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2 years ago
Can someone help me with these questions please?
kotykmax [81]

Step-by-step explanation:

You can solve systems of equations using either substitution or elimination.  For these problems, I recommend elimination.  I'll do the first one as an example.

-3x + 16y = 9

-4x + 8y = 12

Multiply the second equation by -2.

8x − 16y = -24

Add to the first equation (notice the y's cancel out).

(-3x + 16y) + (8x − 16y) = 9 − 24

5x = -15

Solve for x.

x = -3

Now you can plug this into either equation to find y.

-3(-3) + 16y = 9

9 + 16y = 9

y = 0

The solution is (-3, 0).

3 0
3 years ago
Monday 2/3 of the team practiced, Tues 7/8, Wed 1/2,Thurs 3/4. On which day were the most team members present at practice?
MrMuchimi
Tuesday.

1/2 =50%, 3/4=75% 2/3=66.6% and so forth.
8 0
2 years ago
What is the solution to the system of equations? HELP
notsponge [240]
I will solve your system by substitution.<span><span>x=<span>−2</span></span>;<span>y=<span><span><span>23</span>x</span>+3</span></span></span>Step: Solve<span>x=<span>−2</span></span>for x:Step: Substitute<span>−2</span>forxin<span><span>y=<span><span><span>23</span>x</span>+3</span></span>:</span><span>y=<span><span><span>23</span>x</span>+3</span></span><span>y=<span><span><span>23</span><span>(<span>−2</span>)</span></span>+3</span></span><span>y=<span>53</span></span>(Simplify both sides of the equation)

Answer:<span><span>x=<span>−<span><span>2<span> and </span></span>y</span></span></span>=<span>5/3
<span>
so the answer is B (the second choice)
(Hope it helped ^_^)




</span></span></span>
4 0
3 years ago
Read 2 more answers
Describe and correct the error in finding csc θ, given that θ is an acute angle of a right triangle and cos θ =7/11
a_sh-v [17]

Answer:

<h2>cosecθ = 1/sinθ = 11/6√2</h2>

Step-by-step explanation:

Given that  cos θ =7/11, cosec θ = 1/sinθ in trigonometry.

Based on SOH, CAH, TOA;

cosθ = adjacent/hypotenuse = 7/11

adjacent = 7 and hyp = 11

Since sinθ = opp/hyp, we need to get the opposite to be able to calculate sinθ.

Using pythagoras theorem to get the opposite;

hyp^{2} = adj^{2}  + opp ^{2}  \\opp = \sqrt{hyp^{2} - adj^{2}  } \\opp = \sqrt{11^{2} - 7^{2}} \\opp = \sqrt{72} \\opp = 6\sqrt{2}

sinθ = 6√2/11

cosecθ = 1/sinθ = 1/( 6√2/11)

cosecθ = 1/sinθ = 11/6√2

Note the error; cscθ\neq 1/cosθ but cscθ = 1/sinθ

6 0
2 years ago
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