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rodikova [14]
3 years ago
11

According to a recent survey of first-year high school students, 28% chew gum daily. The students were also asked if they had re

cently gotten a cavity filled at the dentist. Of the 47% of first-year students who responded that they had a cavity recently filled, only 39% chewed gum daily. Is chewing gum independent of having a cavity filled recently? Yes, P(Gum) = P(Gum|Cavity). Yes, P(Gum) = P(Cavity|Gum). No, P(Gum) ≠ P(Gum|Cavity). No, P(Gum) ≠ P(Cavity|Gum).
Mathematics
1 answer:
Brut [27]3 years ago
4 0

Answer:

No its not since only 39% if the people who had cavitys had chewed gum regurely

Hope This Helps!!!

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AlladinOne [14]

Answer:

1) See figure attached

2)

a) d = \sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}

And if we replace we got:

d = \sqrt{(10 -6)^2 +(15 -9)^2}= 7.211

b) V = \frac{d}{t}

And if we replace we got:

V = \frac{7.211 ft}{2 sec}=3.606 s

Step-by-step explanation:

Part 1

We can see the plot in the figure attached.

Part 2

a)

For this case we have two points (x_1 , y_1) = (6,9) , (x_2 , y_2) = (10,15)

And we want to find the distance travelled between these two points and we can use the following formula from the euclidian distance between two points:

d = \sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}

And if we replace we got:

d = \sqrt{(10 -6)^2 +(15 -9)^2}= 7.211

b)

Since it takes two seconds in order to go from (6.9) to (10,15) we can use the definition of velocity:

V = \frac{d}{t}

And if we replace we got:

V = \frac{7.211 ft}{2 sec}=3.606 s

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