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OverLord2011 [107]
1 year ago
14

Sabbir arrived at dhaka train station at 9 30 pm write this time using the 24hr clock

Mathematics
2 answers:
mezya [45]1 year ago
6 0

Okay so a 24hr clock is very simple. We usually turn to "PM" when its 12:00 (day time).
In the 24hr clock we do not, we don't use PM or AM but its crucial to know when converting a standard clock time to military time (aka. 24hr).

9:30 PM in 24hr clock is 12:00 + 9:30 because 9:30 is past that first 12 hours.

So in 24hr clock time it is 21:30. You can also count backwards from midnight to figure it out, for example 12:00 AM (midnight), minus 9:30 is 2:30. Therefore 24:00-2:30=21:30.

Hopefully it helps, heart will help me alot too!

SCORPION-xisa [38]1 year ago
4 0

Answer:

21:30

Step-by-step explanation:

The 24hour clock uses 0:00 for midnite. 1am through noon are just 1:00 - 12:00. But the pm hours continue counting up, 13 - 23.

For example, 1pm is 13:00 and 2pm is 14:00.

To change to 24 hour clock, for pm hours, ADD 12.

So for your problem, 9:30 is:

9:30 + 12

= 21:30

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Suppose that 30% of all houses need a paint job. Also, 15% of all houses need both a paint job and a new roof. Further, 7% of al
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0.0105

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given that 30% of all houses need a paint job. Also, 15% of all houses need both a paint job and a new roof. Further, 7% of all houses that need a new roof also need new windows.

If total houses are 100, then 30 require a paint along.  15 require both paint and new roof.

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8 0
3 years ago
Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the cir
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To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.

If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle.  If the left-hand side is equal to r^2, the point is ON the circle.
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For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
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A. (-1,1), 
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B. (10,0)
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C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49  [outside circle]


5 0
3 years ago
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