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3 years ago

50 POINTS!!! Please give an explanation and answer to this word problem. I am not just giving you 50 points for an answer!

Mathematics

2 answers:

denpristay [2]3 years ago

7 0

Answer:

34 miles

Step-by-step explanation:

The distance AB is 200 miles

Let them meet after x hours, we have equation below:

17x + 83x = 200
100x = 200
x = 2 hours

The cyclist will drive the distance from point A:

17*2 = 34 miles

They will meet at 34 miles from point A

valentinak56 [21]3 years ago

5 0

Answer:

D from A = 34 miles

Step-by-step explanation:

We know the total distance is 200 miles (this would be when they meet)

The rate for cyclist A is 17

The rate for cyclist B is 83

We know the time is equal

Distance = rate * time

Distance for A = 17t

Distance for B = 83 t

The total distance = distance for A + distance for B

200 = 17t + 83t

Combine like terms

200 = 100t

Divide each side by t

200/100 = 100t/100

They will meet at t =2 hours

I know you say we need the distance

Distance for A

D = 17*t

We now know t

D = 17*2

D = 34 miles

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3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago