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jonny [76]
3 years ago
15

I'M GIVING OUT BRAINLIEST! Terry and Tao are canoeing on a lake that is 2350 meters across from the west side of the lake to the

east side. Terry rows at 50 meters per minute and Tao rows at 10 meters per minute. Terry starts rowing at 2 PM from the west end of the lake, and Tao starts rowing from the east end of the lake at 2:05 PM. If they always row directly towards each other, at what time will the two meet?
Mathematics
1 answer:
OLEGan [10]3 years ago
6 0

Answer:

  2:40 PM

Step-by-step explanation:

Terry's 5-minute head start has reduced the distance between them by ...

  (5 min)(50 m/min) = 250 m

After Tao starts, at 2:05 PM, the remaining 2350 -250 = 2100 m of distance is being closed at the rate of 50 +10 = 60 m/min. It will take ...

  (2100 m)/(60 m/min) = 35 min

to close the remaining distance between the rowers. 35 minutes after Tao starts is 2:05 + 0:35 = 2:40 PM.

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Answer:

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Step-by-step explanation:

an integer is a natural counting number, such as 1...2...3...4... and so on....

zero is not considered an integer because it is a neutral number.

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3 years ago
Find the equation of the line containing the points (20,-7) and (10,-2). Type your answer in this order and don't use spaces or
aliina [53]

(\stackrel{x_1}{20}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-2}-\stackrel{y1}{(-7)}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{20}}}\implies \cfrac{-2+7}{-10}\implies -\cfrac{1}{2}

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-7)}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{20}) \\\\\\ y+7=-\cfrac{1}{2}x+10\implies y=-\cfrac{1}{2}x+3

4 0
2 years ago
According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e
slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

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.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that p = 0.09

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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

P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225

P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827

P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082

P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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