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iVinArrow [24]
3 years ago
11

A fair number cube is rolled. what are the odds in favor of rolling a 4

Mathematics
2 answers:
morpeh [17]3 years ago
7 0
There are 6 sides in a cube.
Rolling one number, the four, out of 6 numbers is 1/6.
The odds are 1/6, or 1 in 6
irinina [24]3 years ago
3 0
Number of possibilities for 1 roll = 6. Probability of rolling a 4 = 1/6. . Odds are 5 to 1 against it. 1 to 5 in favor.
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To get this you convert the km into cm as it is easier and then it becomes something like this 2cm : 250000cm
then divide or half to get to the simpler form like this 2cm divided by 2=1cm and 250000cm divided by 2= 125000cm 
then put into ratio is 1:125000

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Consider that an = 4n-1 represents a sequence. What is the tenth term? A) 361 B) 512 C) 1,024 D) 262,144
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The Answer is C. Hopefully this will help you out.
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Charles saves $y. Sharon saves 3 times as much as Charles. Jason saves 50.00 more thN Sharon. How much do they save
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If f(x)=x^1/2-x and g(x)=2x^3-x^1/2-x, find f(x)-g(x)
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Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north
Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

  • x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

  • Child 1 final position => x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t
  • Child 2 final position => x_{2f} =0+v_{2} t

4) Sum the equations and equate to 3

  • x_{1f} +x_{2f} =3

5) Substitute the values we already know

  • 1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

7 0
3 years ago
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