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vovangra [49]
3 years ago
13

How many rows are needed to seat all students. When there are 846 students and each row holds 18 people

Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

The answer is 47 rows.

Step-by-step explanation:

If we have 846 students, and each row can hold 18 people.

We just divide the total students into the number of people sitting by row.

846 ÷ 18 = 47.

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When we rotate clockwise or counterclockwise, the two rotations should always add up to ( ) degrees
oksian1 [2.3K]

Answer:

360 degrees

Step-by-step explanation:

6 0
3 years ago
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A five-sided solid has the numbers 1, 2, 3, 4, and 5. What is the probability of rolling two five-sided number solids and gettin
Klio2033 [76]
We assume the probability on each side is equally probable with probability 1/5.
sum=4 has outcomes:{1,4; 2,3; 3,2; 4,1}  4 possible outcomes
sum=8 has outcomes:{3,5; 4,4; 5,3} 3 possible outcomes.
Total possible outcomes = 5*5=25
there probability of rolling a sum of 4 or 8, by the law of addition, equals
4/25+3/25=7/25

Note: a regular (i.e. fully symmetric) five-sided solid does not exist, so there has to be asymmetry among the probabilities of the five possible outcomes.  In addition, it does not have a "top" face, so that makes rolling a five-sided solid a little more difficult to visualize.
7 0
3 years ago
Change the line 3x − 8y = 5 into slope-intercept form.
Nimfa-mama [501]
Slope is y=mx+b
m=slope
b=y intercept

solve for y basically


3x-8y=5
minus 3x both sides
-8y=-3x+5
divide both sides by -8
y=\frac{3}{8}x-\frac{5}{8}
4 0
3 years ago
1 > -1/2x + 5 solve for x
RUDIKE [14]

Answer:

x is greater than 8

Step-by-step explanation:

6 0
2 years ago
If the distance covered by an object in time t is given by s(t)=t^2+5t , where s(t) is in meters and t is in seconds, what is th
alisha [4.7K]

Answer:

E. 44 meters

Step-by-step explanation:

The function that models the distance covered by the object is

s(t)=t^2+5t

where s(t) is in meters and t is in seconds.

The distance covered by the object after 1 second is

s(1)=1^2+5(1)=6m

The distance covered by the object after 1 second is

s(1)=5^2+5(5)=50m

The distance covered between 1 second and 5 seconds is

50-6=44m

6 0
3 years ago
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