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hodyreva [135]
3 years ago
6

48 kilometers in 3 hours =

Mathematics
2 answers:
Genrish500 [490]3 years ago
6 0
16 kilometer an hour
lana [24]3 years ago
3 0
U think its 16 kilometer in hour
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What is three ratios that are equivalent to 18:4
musickatia [10]
We can make ratios equivalent to 18:4 by dividing and multiplying on it.
1. (18:4)/2=9:2
2. (18:4)*2=36:8
3. (18:4)*3=54:12

You're welcome :).
5 0
3 years ago
Read 2 more answers
The probability a bus will be full is 65% which words or words best describes the likelihood the bus will be full?
allsm [11]
Due to the fact that the question states there is a 65% chance the bus will be full, that is a very high probability. Therefore, there is only a 35% chance it wont be full. Or you can say it in fractional form: 65/100 chance of it being full, and a 35/100 chance of it being less than full.<span />
8 0
4 years ago
Pls help it due in a hour
Levart [38]

Answer:

The coordinates of the Midpoint of AB will be: (1/2, 5)

Step-by-step explanation:

Given the points

  • A(-2, 6)
  • B(3, 4)

Finding the midpoint between (-2, 6) and (3, 4)

\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

\left(x_1,\:y_1\right)=\left(-2,\:6\right),\:\left(x_2,\:y_2\right)=\left(3,\:4\right)

M.P=\left(\frac{3-2}{2},\:\frac{4+6}{2}\right)

        =\left(\frac{1}{2},\:5\right)

Therefore, the coordinates of the Midpoint of AB will be: (1/2, 5)

6 0
3 years ago
Each week Peggy drives two routes, route X and route Y. One week she drives route X three times and route Y twice. she drives a
navik [9.2K]
Ok, great to have you! I will finally answer your question.
This seems to be a system of equations. I solve most of mine on Desmos.
Let us write two equations to model this:
3x+2y=144
2x+5y =217
We get the graph. Note the intersection at (26,33).
That means that route X is 26 miles long,and route Y is 33 miles long!.
Hope this helps.

4 0
3 years ago
You are trying to figure out how many gumballs you need to fill a 7.3 x 5.0 x 9.4 rectangular box for Halloween. Each gumball ha
riadik2000 [5.3K]

According to the calculations made, 410 gumballs will be needed to fill the box.

Since you are trying to figure out how many gumballs you need to fill a 7.3 x 5.0 x 9.4 rectangular box for Halloween, and each gumball has a radius of 1/2 in, if the packing density for spheres is 5/8 of the volume will be filled with gumballs while the rest will be air how many gumballs will be needed, to determine this amount the following calculation must be performed:

  • (Volume of box x 5/8) / volume of gumballs = Amount of gumballs  
  • Volume of a sphere = 4/3 x 3.14 x (radius x radius x radius)  
  • Volume of a gumball = 4/3 x 3.14 x (0.5 x 0.5 x 0.5) = 0.5235 inches  
  • ((7.3 x 5 x 9.4) x 5/8) / 0.523 = X
  • (343.1 x 5/8) / 0.523 = X
  • 214.4375 / 0.523 = X
  • 410 = X

Therefore, 410 gumballs will be needed to fill the box.

Learn more in brainly.com/question/1578538

4 0
3 years ago
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