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valina [46]
3 years ago
9

There are 620 students and 31 teachers at the Turner Middle School.There are 950 students and 38 teachers at the Green Middle Sc

hool.
a. Find the ratios of students to teachers at each school. Can you form a proportion with these ratios ? Why or why not?

b. Suppose Green school wants to make its student-teacher ratio close to that at Turner school. How many new teachers would Green have to hire?

c. Can you write a part-to-whole ratio using the data provided? Explain.
Mathematics
2 answers:
kicyunya [14]3 years ago
8 0
We begin with the <span> Turner Middle School:
Ratio of students to teachers: </span>\frac{620}{31}
Green Middle School:
Ratio of students to teachers: \frac{950}{38}
Let check that we cannot form a proportion:
\frac{620}{31}=20\text{ and } \frac{950}{38}=25
The two ratios are not equals so we cannot form a proportion. 
Let x be the number of <span>teachers that Green have to hire in order to make the two ratios equal. We get the equation:
</span>\frac{950}{38+x}=\frac{620}{31}\text{ and we solve the equation like this:}\\38+x=\frac{950\times31}{620}=47.5 \text{ and then:}\\x=47.5-38=9.5

<span>The Green school need to hire 9 teachers. 

Part-to-whole ratio:
</span>Turner Middle School:
 \frac{620}{31+620}
Green Middle School:
\frac{950}{38+950}
suter [353]3 years ago
8 0
Can someone help with the answer please
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4 0
3 years ago
What is a correct congruence statement for the triangles shown?<br> Enter your answer in the box.
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Answer:

ASA

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6 0
1 year ago
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Question Down below.
expeople1 [14]

Answer:

there equal

Step-by-step explanation:

the absolute value of 3 is 3

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hope this helps

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Have a great day!

8 0
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zhuklara [117]
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4 0
3 years ago
PLEASE HELP! I'll give 5 out of 5 stars, give thanks, and give as many points as I can.
Liula [17]

Answer:

\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}

Step-by-step explanation:

\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)

\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)

7 0
3 years ago
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