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ValentinkaMS [17]
2 years ago
7

7. for normal situations, the minimum speed limit throughout the u.s. interstate highway system is 72 km/hr which is 45 miles pe

r hour what is the distance, in kilometers, traveled by a car moving at this speed for 3.3 hours
8. the maximum posted speed limit on the u.s. interstate highway system may be found in rural areas of several Western States this maximum speed is 121 kilometers per hour (75 miles per hour) what is the distance in kilometers traveled by a car moving at the speed for 3.3 hours

Mathematics
1 answer:
gladu [14]2 years ago
3 0

Answer:

Question 7: Distance = 237.6 km

Question 8: Distance = 399.3 km

Step-by-step explanation:

Question 7:

Given:

Minimum speed (s) = 72 km/hr

Time taken (t) = 3.3 hours.

We know that, distance traveled is equal to the product of speed and time.

Therefore using distance formula, we get:

Distance = Minimum speed × Time

Distance = s\times t

Distance = 72\times 3.3 = 237.6\ km

Therefore, distance traveled at this speed is 237.6 km.

Question 8:

Maximum speed (S) = 121 km/hr

Time taken (T) = 3.3 hours.

We know that, distance traveled is equal to the product of speed and time.

Therefore using distance formula, we get:

Distance = Maximum speed × Time

Distance = S\times T

Distance = 121\times 3.3 = 399.3\ km

Therefore, distance traveled at this speed is 399.3 km.

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Step-by-step explanation:

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assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the
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65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value x_{0} such that P(X < x_{0}) = 0.76, but, 0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5), i.e., (x_{0}-63.6)/2.5 is the 76th percentile of the standard normal distribution. So, (x_{0}-63.6)/2.5 = 0.7063, x_{0} =63.6+(2.5)(0.7063) = 65.3658. Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.

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3 years ago
Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili
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Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}

Use the Bayes' theorem to compute the value of P (B | D) as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

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Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

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All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

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Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

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