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stealth61 [152]
3 years ago
14

Two cars at the same point on I-75. One heads north going 75mph and one heads south going 60mph. How long will it take for the c

ars to be at 420 miles apart?
Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

It will take 3 hours 6 minutes 40 seconds for the cars to be 420 miles apart.

Step-by-step explanation:

Step 1; The cars are traveling in the opposite direction at different speeds. One is going north at a speed of 75 mph while the other is going 60 mph. So for every hour, the cars are traveling they increase the distance between in between themselves by 75 miles + 60 miles = 135 miles.

Step 2; Assume that in x hours the distance between them is 420 miles. To calculate x we divide the distance to be traveled by the distance being traveled every hour.

x = distance to be covered / distance being traveled every hour

= 420 / 135 = 3.11 hours

We multiply the 0.11 hours with 60 to convert it into minutes. 0.11 × 60 = 6.66 minutes and if we do the same for seconds, 0.66 minutes × 60 = 40 seconds.

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A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\

A. P(x=6)

P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\

B. P(x≥10)

P(x\geq10)=1-P(x

P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\

P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\

P(x\geq10)=1-(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828+0.1654+0.1241+0.0781)\\\\P(x\geq10)=1-0.9287=0.0713

C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:

E(x)=\mu=n\cdot p=25\cdot 0.25=6.25\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot 0.25\cdot 0.75}=\sqrt{4.69}\approx2.17

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