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-BARSIC- [3]
3 years ago
9

there are 150 more passenger cars than trucks on the highway. for every 10 passenger cars there are 7 trucks. how many trucks ar

e on the highway?
Mathematics
2 answers:
Hoochie [10]3 years ago
8 0

Answer:

There would be 500 passenger cars and 350 trucks.

Step-by-step explanation:

In order to find this, we need to create a system of equations. Start by making the number of passenger cars as x and the number of trucks at y. Now the first equation can be the difference of the two.

x - y = 150

The second equation can make use of the proportion. If we multiply by the opposites we'll get the difference of 0.

7x - 10y = 0

Now we multiply the first equation by -10 and add together to solve for x.

-10x + 10y = -1500

7x - 10y = 0

-------------------

-3x = -1500

x = 500

Now we can find the amount of trucks by using either equation.

x - y = 150

500 - y = 150

-y = -350

y = 350

lora16 [44]3 years ago
5 0

Answer:

500

Step-by-step explanation:

Given: P = 150 - T where P is passenger cars and T is trucks

10P = 7T ⇔ P = \frac{7T}{10}

Plug in equation 2 into equation 1:

\frac{7T}{10} = 150 - T

Solving for T, we get 500

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Answer:

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

Step-by-step explanation:

For each U.S. resident, there are only two outcomes possible. Either they have blood type AB, or they do not. This means that we can solve this problem using binomial probability distribution concepts.

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}.\pi^{x}.(1-\pi)^{n-x}

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

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

And \pi is the probability of X happening.

In this problem, we have that:

50 U.S residents are sampled, so n = 50

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P(X = 2) = C_{50,2}.(0.04)^{2}.(0.96)^{48} = 0.2762

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

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I wrote it down better
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