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Snowcat [4.5K]
3 years ago
14

5. Is the equation 2x + 8 = y a direct variation? (yes or no)

Mathematics
1 answer:
aalyn [17]3 years ago
8 0

Answer:

Yes.It is

Step-by-step explanation:

A relationship between two variables in which one is a constant multiple of the other. the types of variation include direct, inverse, joint, and combined variation keep that in hand

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What is the distance between (-4,6) and (3,7)
klasskru [66]

Answer:

\sqrt{50} or 7.0710678118654755

Step-by-step explanation:

The distance formula is d=\sqrt{(x_{2}-x_{1})^{2}- (y_{2}-y_{1})^{2} }.

We then want to fill it out wiht our numbers.

d=\sqrt{(3--4)^{2}- (7-6)^{2} }


3 0
4 years ago
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Artyom0805 [142]

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(y2-y1)/(x2-x1)

and simplifying

6 0
3 years ago
Solve this question please
mr_godi [17]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
The three circles are all centered at the center of the board and are of radii 1, 2, and 3, respectively.Darts landing within th
FromTheMoon [43]

Answer:

a) P=0.262

b) P=0.349

c) P=0.215

d) E(x)=12.217

e) P=0.603

Step-by-step explanation:

The question is incomplete.

Complete question: "The three circles are all centered at the center of the board (square of side 6) and are of radii 1, 2, and 3, respectively.Darts landing within the circle of radius 1 score 30 points, those landing outside this circle, butwithin the circle of radius 2, are worth 20 points, and those landing outside the circle of radius2, but within the circle of radius 3, are worth 10 points. Darts that do not land within the circleof radius 3 do not score any points. Assume that each dart that you throw will land on a point uniformly distributed in the square, find the probabilities of the accompanying events"

(a) You score 20 on a throw of the dart.

(b) You score at least 20 on a throw of a dart.

(c) You score 0 on a throw of a dart.

(d) The expected value on a throw of a dart.

(e) Both of your first two throws score at least 10.

(f) Your total score after two throws is 30.

As the probabilities are uniformly distributed within the area of the board, the probabilities are proportional to the area occupied by the segment.

(a) To score 20 in one throw, the probabilities are

P(x=20)=P(x=20\&30)-P(x=30)=\frac{\pi r_2^2}{L^2} -\frac{\pi r_1^2}{L^2}\\\\P(x=20)=\frac{\pi(r_2^2-r_1^2)}{L^2}=\frac{3.14*(2^2-1^2)}{6^2}=\frac{3.14*3}{36} =0.262

(b) To score at least 20 in one throw, the probabilities are:

P(x\geq 20)=P(x=20\&30)=\frac{\pi r_2^2}{L^2}=\frac{3.14*2^2}{6^2} =0.349

(c) To score 0 in one throw, the probabilities are:

P(x=0)=1-P(x>0)=1-\frac{\pi r_3^2}{L^2} =1-\frac{3.14*3^2}{6^2} =1-0.785=0.215

(d) Expected value

E(x)=P(0)*0+P(10)*10+P(20)*20+P(30)*30\\\\E(x)=0+\frac{\pi(r_3^2-r_2^2)}{L^2}*10+ \frac{\pi(r_2^2-r_1^2)}{L^2}*20+\frac{\pi(r_1^2)}{L^2}*30\\\\E(x)=\pi[\frac{(3^2-2^2)}{6^2}*10+\frac{(2^2-1^2)}{6^2}*20+\frac{1^2}{6^2}*30]\\\\E(x)=\pi[1.389+1.667+0.833]=3.889\pi=12.217

(e) Both of the first throws score at least 10:

P(x_1\geq 10; x_2\geq 10)=P(x\geq 10)^2=(\frac{\pi r_3^2}{L^2} )^2=(\frac{3.14*3^2}{6^2} )^2=0.785^2=0.616

(f) Your total score after two throws is 30.

This can happen as:

1- 1st score: 30, 2nd score: 0.

2- 1st score: 0, 2nd score: 30.

3- 1st score: 10, 2nd score: 20.

4- 1st score: 20, 2nd score: 10.

1 and 2 have the same probability, as do 3 and 4, so we can add them.

P(2x=30)=2*P(x_1=30;x_2=0)+2*P(x_1=20;x_2=10)\\\\P(2x=30)=2*P(x_1=30)P(x_2=0)+2*P(x_1=20)P(x_2=10)\\\\P(2x=30)=2*\frac{\pi r_1^2}{L^2}*(1-\frac{\pi r_3^2}{L^2})+2*\frac{\pi(r_2^2-r_1^2)}{L^2}*\frac{\pi(r_3^2-r_2^2)}{L^2}\\\\P(2x=30)=2*\frac{\pi*1^2}{6^2}*(1-\frac{\pi*3^2}{6^2})+2*\frac{\pi(2^2-1^2)}{6^2}*\frac{\pi(3^2-2^2)}{6^2}\\\\P(2x=30)=2*0.872*0.215+2*0.262*0.436=0.375+0.228=0.603

6 0
3 years ago
What is the value of h when the function is converted to vertex form?
Alenkinab [10]

Answer: H=3

                    ^ I noticed you need help on this one the answer is very simple




7 0
3 years ago
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