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joja [24]
3 years ago
9

For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation.

Mathematics
1 answer:
Rashid [163]3 years ago
5 0
Our equation for this table could be an equation in slope-intercept form:
y=mx+b
m=slope of the line:
b=y-intercept.

1) we have to calculate the slope of this line.
We need two points, you can choose two points of this table, for example:
the points (4,6) and (10,15)

m=(y-y₀)/(x-x₀)
m=(15-6)/(10-4)
m=9/6
m=3/2

2) we have to find "b";
We have this equation:
y=3/2 x +b

Now, we choose any point of this table; For example (8,12), now we find "b",
y=3/2 x+b
12=3/2 (8)+b
12=12+b
b=12-12
b=0

Therefore: the equation would be y=3/2 x 

More easy:
if y vary directly with x, then: y=(y₂-y₁)/(x₂-x₁) x
You choose any two points of this table:
(8,12)
(10,15)

y=(15-12)/(10-8) x
y=3/2 x

Answer: y=3/2 x
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Answer:

1.B

2.C

3.D

4.

a)1/9

b)5/12

5.2/5

Step-by-step explanation:

1.

Since there are 15 total outcomes of which 6 are desirable, the probability of randomly getting a red ball is 5/16 or choice B.

2.

Since there are a total of 8+7+6=21 outcomes, out of which 7 are neither red nor green, there is a 7/21=1/3 probability of that outcome, or answer choice C.

3.

Out of the 52 possible outcomes, 12 are face cards. Therefore, the probability of drawing a face card is 12/52=3/13, or answer choice D.

4.

a)

For each die, there are a total of 6 possible outcomes. This means that in total there are 6*6=36 possible combinations. Now, let's do some casework.

If the first die is a 3, then the second die must be a 6. If the first die is a 4, the second die must be a 5. If the first die is a 5, then the second die must be a 4. And if the first die is a 6, then the second die must be a 3. You cannot have any numbers less than 3, because then you would not be able to make the number 9. Therefore, there is a 4/36=1/9 chance of getting a sum of 9.

b)

All of the possible prime numbers that can be formed with the dice are 2, 3, 5, 7, and 11. For 2, there is only one possible situation where this can happen. For 3, there are 2 possible. For 5, there is 4-1, 3-2, 2-3, and 1-4, or 4 possibilities. For 7, there is 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1, or 6 total possibilities. Finally, for 11, there is only 5-6 and 6-5, or 2 possible. In total, the probability of getting a prime sum is 15/36=5/12.

5.

The only numbers which are a multiple of 3 or 5 from 1 to 20 are: 3, 5, 9, 10, 12, 15, 18, and 20. This means that out of the 20 total possible options, 8 of them are a multiple of 3 or 5, which is a total chance of 2/5.

Hope this helps!

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3 years ago
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