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1 year ago

The probability that a vehicle entering the Luray

1 answer:

4 0

Probability that a vehicle entering the Luray Caverns has Canadian license plates = 0.12
Probability that it is a camper = 0.28
Probability that it is a camper with Canadian license plates = 0.09

a. Probability that a camper entering the Luray Caverns has Canadian
license plates = 0.09

b. Probability that a vehicle with Canadian
license plates entering the Luray Caverns is a camper = 0.09/0.12
= 0.75

c. Probability that a vehicle entering the Luray Caverns
does not have Canadian plates or is not a camper = 0.88 + 0.72 - 0.81
= 0.80

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HELP!!<br> What is the equation of the circle shown in the graph?

ale4655 [162]

9514 1404 393

Answer:

(x +6)^2 +(y -4)^2 = 36

Step-by-step explanation:

The center is (-6, 4) and the radius is 6. Putting those into the standard form equation, you have ...

(x -h)^2 +(y -k)^2 = r^2 . . . . . . center (h, k), radius r

(x -(-6))^2 +(y -4)^2 = 6^2 . . . . numbers filled in

(x +6)^2 +(y -4)^2 = 36 . . . . . . cleaned up a bit

8 0

1 year ago

The given equation has been solved in the table. In which step was the subtraction property of equality applied?

atroni [7]

Answer:

Option (D)

Step-by-step explanation:

Subtraction property of equality tells that whatever subtracted from one side of the equation must be subtracted from the other side.

If x + 2 = 2,

By the property of subtraction of equality,

x + 2 - 2 = 2 - 2

x = 0

But in the given question,

$\frac{x}{2}-7=-7$

$\frac{x}{2}-7+7=-7+7$

shows the addition property of equality in step (2)

Therefore, subtraction property of equality was not applied.

Option (D) will be the answer.

5 0

1 year ago

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago

The number of airline passengers in 1990 was 466 million. The number of passengers traveling by airplane each year has increased

taurus [48]

Answer:

2010.

Step-by-step explanation:

We have been given an exponential growth formula $P(t)=466\cdot 1.035^t$ , which represents number of passengers traveling by airplane since 1990.

To find the year in which 900 million passengers will travel by airline, we will equate the given formula by 900 and solve for t as:

$900=466\cdot 1.035^t$

$\frac{900}{466}=\frac{466\cdot 1.035^t}{466}$ $1.9313304721030043=1.035^t$

Take natural log of both sides:

$\text{ln}(1.9313304721030043)=\text{ln}(1.035^t)$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$\text{ln}(1.9313304721030043)=t\cdot \text{ln}(1.035)$

$0.658209129198=t\cdot 0.034401426717$

$\frac{0.658209129198}{0.034401426717}=\frac{t\cdot 0.034401426717}{0.034401426717}$

$19.1331927775=t\\\\t=19.1331927775$

This means that in the 20th year since 1990, 900 million passengers would travel by airline.

$1990+20=2010$

Therefore, 900 million passengers would travel by airline in 2010.

6 0

1 year ago

Find the gradients of lines A and B

mel-nik [20]

Step-by-step explanation:

$the \: gradient \: of \: line \: a \: is$

$m = \frac{4}{1 } \\ m = 4$

$the \: gradient \: of \: line \: b \: is$

$m = \frac{ - 2}{1} \\ m = - 2$

6 0

2 years ago