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Anika [276]
3 years ago
11

Someone lives in the Northeast region and your best friend's family lives in the Midwest region so you decide to compare ratings

from those two regions. Make a graph that shows the importance of recycling according to region.
“How important is recycling garbage? Rate your answer on the scale from 0 (not important) to 1000 (very important).”
Data:
South 800
Midwest 600
South 1000
Northeast 914
Midwest 941
and MUCH MORE

QUESTION:
What type of graph would be correct? A bar graph, histogram, a pie graph? Explain why.
Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

a bar graph

Step-by-step explanation:

I think donno if it is right.

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Melinda spends $5 at each of the 14<br> stores she visits. How much money did she spend ?
Paha777 [63]

Answer:

$70

Step-by-step explanation:

5 dollars * 14 stores= $70

4 0
3 years ago
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I don’t understand how to factor fractions :( can someone help me with this problem
anzhelika [568]
Can you take a picture of the whole problem? i may be able to solve it then. Thanks!
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How do you rewrite quadratic equations
disa [49]
A quadratic should be in the form ax²+bx+c=0
where a b and c are all numbers 
8 0
3 years ago
In TUV below, find m T to the nearest degree
Andre45 [30]

Answer:

T = 88 degrees to the nearest degree

Step-by-step explanation:

To find the measure of the angle T, we can use the cosine rule

We have the formula as;

t^2 = u^2 + v^2 - 2uv Cos T

t = 8.1

u = 7.1

v = 4.2

8.1^2 = 7.1^2 + 4.2^2 - 2(7.1)(4.2) cos T

-2.44 = -59.64 cos T

cos T = (2.44)/(59.64)

T = arc cos (2.44/59.64)

T = 87.66

to the nearest degree, this is 88 degrees

7 0
3 years ago
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Determine the current through each of the LEDs in the circuits below. Which LED will be
Anika [276]

a. I = 6. 1 × 10^-4 A

b. I = 2. 6 × 10^-3 A

c. I = 0. 04 A

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

<h3>How to determine the current</h3>

The formula for finding current

I = V/R

Where v = voltage

R = resistance

A. V = 12V

R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series

I = \frac{12}{19700}

I = 6. 1 × 10^-4 A

B. V = 9V

R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series

I = \frac{12}{4700}

I = 2. 6 × 10^-3 A

C.  V=  12V

1/R = \frac{1}{750} + \frac{1}{1200} + \frac{1}{950 } = 3. 22 × 10^-3

R = \frac{1}{3. 22 * 10 ^-3} = 310. 56 Ω

I = \frac{12}{310. 56}

I = 0. 04 A

It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

Learn more about Ohms law here:

brainly.com/question/14296509

#SPJ1

8 0
1 year ago
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