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xxMikexx [17]
3 years ago
7

A car and a scooter are moving in opposite directions and cross each other at a common point. if the average speed of the car is

50 miles per hour and that of the scooter is 40 miles per hour,after how many hours will they be 180 miles apart?
Mathematics
1 answer:
Tema [17]3 years ago
6 0
As they are traveling in opposite directions you need to do 50 + 40 to find out how far apart they travel every hour. 50 + 40 = 90. So you know that they are 90 miles apart every hour.
180 / 90 = 2. so you know that it takes them 2 hours to be 180 miles apart.
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Ivahew [28]
The price for each ticket is 22.41
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3 years ago
A detachment of n soldiers must cross a wide and deep river with no bridge in sight. They notice two 12- year-old boys playing i
Vitek1552 [10]

Answer:

Steps 1 - 4 below explains it.

Step-by-step explanation:

The way that the soldiers get across the river and leave the boys in joint possession of the boat is;

1) The two 12- year - old boys will take the boat to the other side

2) Thereafter, one of the 12- year - old boys will return with the boat.

3) A soldier will now take the boat to the other side and the soldier will remain there while the other boy returns the boat.

4) Step 1 - 3 is a total of 4 trips. Thus, trip would be repeated the total of n times and the n number of soldiers would get across the river and leave the boys in joint possession of the boat after the total of 4n trips.

5 0
3 years ago
The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

We have  

\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

hence,

\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

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Anna007 [38]

Answer:

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Step-by-step explanation:

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3 years ago
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Softa [21]

Answer:The approximate square root is 2.23

Step-by-step explanation:√ 5  = 2.23

Thus:    2.23  x  2.23 = 5

6 0
3 years ago
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