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Marta_Voda [28]
3 years ago
10

A graph titled Federal Minimum Wage Results, from 1970 to 2005 every 5 years has years on the x-axis and minimum wage (dollars)

on the y-axis. Points are at (1, 1.5), (2, 2), (3, 3), (4, 3.2), (5, 3.8), (6, 4.1), (7, 5), (8, 5).
The scatterplot shows the federal minimum wage rates in the United States, calculated every five years from 1970 (year 1 on the graph) to 2005 (year 8).

Which is the best equation of the trend line for the scatterplot?
y = –0.45x + 1.55
y = –0.45x – 1.55
y = 0.45x + 1.55
y = 0.45x – 1.55
Mathematics
1 answer:
Novay_Z [31]3 years ago
4 0

Answer:

it’s c on edg2020

Step-by-step explanation:

took the test

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I believe the answer is 13 ways to make 27 cents using Quarters, dimes, nickels, and pennies.

Step-by-step explanation:

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What is y=7/3x+5?<br> x=-3
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Hello!

Ok so we have y= 7/3x + 5 and initial condition of x=-3 so we just plug -3 in for x and chug along at this point! :)

y=(7/3)*(-3/1)+5= -7+5 = -2

Hope this helps! Any questions please just ask! Thank you!!
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It takes 52 minutes for 5 people to paint 5 walls. How many minutes does it take 20 people to paint 20 walls?
balu736 [363]
It  will take 20 people to paint 5 walls in 52/4 = 13 minutes 
<span>so time  for 20 people to paint 20 walls = 4 * 13 = 52 minutes</span><span>



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4 0
4 years ago
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Please help! The last answer I need!
ruslelena [56]

Answer

I would choose the answer c

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6 0
3 years ago
It’s a new semester!students are grouped into three clubs, which each has 10,4 and 5 students.In how many ways can teacher sleep
My name is Ann [436]

The total number of ways to select 2 students from the 3 clubs is 55 ways.

Since there are three clubs which each has 10, 4 and 5 students, and we require the number of ways a teacher can select 2 students so that they are from different clubs.

Since order doesn't matter and anybody can be chosen first, we use combination theory.

<h3 /><h3>Number of ways of selecting from the first two clubs</h3>

Since we have two slots, to select the first person from the first club, we ¹⁰C₁. For the second student from the club of 4, we have ⁴C₁. Also, there are 2 ways of selecting the two students.

So, there are ¹⁰C₁ × ⁴C₁/2!

= 10 × 4/2

= 20 ways from the first two clubs.

<h3 /><h3>Number of ways of selecting from the next two clubs</h3>

For the next two clubs of 4 and 5 students, for the first slot, we have ⁴C₁. For the second student, we have ⁵C₁. Also, there are 2 ways of selecting the two students.

So, there are ⁴C₁ × ⁵C₁/2!

= 4 × 5/2

= 10 ways from the next two clubs.

<h3 /><h3>Number of ways of selecting from the last two clubs</h3>

For the first and last club of 10 and 5 students, for the first slot, we have ¹⁰C₁. For the second student, we have ⁵C₁.  Also, there are 2 ways of selecting the two students.

So, there are ¹⁰P₁ × ⁵P₁/2!

= 10 × 5/2

= 25 ways from the first and last club.

<h3 /><h3>Total number of ways of selecting two students from the 3 clubs.</h3>

So, the total number of ways to select 2 students from the 3 clubs is 20 + 10 + 25 = 55 ways.

So, the total number of ways to select 2 students from the 3 clubs is 55 ways.

Learn more about combinations here:

brainly.com/question/25990169

7 0
3 years ago
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