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JulsSmile [24]
3 years ago
7

Forty five percent of students said they would go again to see the movie. If there are 60 students, how many would want to go ag

ain
Mathematics
1 answer:
Anestetic [448]3 years ago
5 0
27 students would want to go again since 45% of 60 is 27
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Can someone help me with this please? I will mark you brainliest
lina2011 [118]

Answer:

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Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

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Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

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Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.vvOn the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.vcOn the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. ...

Draw a bar extending from the lower value of each interval to the lower value of the next interval.

Step-by-step explanation:

6 0
3 years ago
If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?
fgiga [73]
X in the first equation

because 3x + 6y = 9 can be reduced by dividing by 3, thus, giving u 
x + 2y = 3.....x = -2y + 3...which u would sub in for x in the other equation
7 0
3 years ago
Ok the first question confuses me what is the volume of the prisom show in the top?
Veronika [31]
Split the shape in to two this is the rectangular side.
15+6=21
21×7=147
147×4=588in
square shape
6+6=12
12×12=144
144×4=576in
then add the answers together
588+576=1164in
1164 is how many more inches greater
5 0
4 years ago
Can you help with both questions 4 and 5?
cupoosta [38]
Well since the triangle is split into 2 triangles and they both are the same size so BD on the bottom is 15 which means naturally BC must be 30. Five I'm not sure about
8 0
3 years ago
Read 2 more answers
ustin wants to use 188 ft of fencing to fence off the greatest possible rectangular area for a garden. What dimensions should he
Tema [17]

Answer:

I would say that it should be 48 ft long and 46 ft wide, the area would be 2208 square feet

Step-by-step explanation:

I will try my best to answer this.

I would start by dividing 188 by 4

this gives you a 47*47 square

if it needs to be a rectangle I would then make the sides 48, 48, 46, and 46

48*46 would be the dimensions

the area would be 2208 square feet

5 0
4 years ago
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