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mr Goodwill [35]
3 years ago
8

A tire manufacturer estimates that their tires last, an average of 40,000 miles, with a standard deviation of 5000. What is the

probability that a randomly chosen tire from this manufacturer lasts between 30,000 miles and 50,000 miles? (Give your answer as a decimal rounded to 4 decimal places.) check Reference Answer: 2. The daily high temperature on October 31 in a certain city is normally distributed with 50 and 0 8. What is the probability that the high temperature on October 31 in a randomly chosen year will be between 46 degrees and 58 degrees? (Give your answer as a decimal rounded to 4 decimal places.)
Mathematics
1 answer:
Oxana [17]3 years ago
3 0

Answer:

0.9554,0.8188

Step-by-step explanation:

Given that a tire manufacturer estimates that their tires last, an average of 40,000 miles, with a standard deviation of 5000.

X - duration of tires is N(40000,5000)

Or Z = \frac{x-40000}{5000} is N(0,1)

the probability that a randomly chosen tire from this manufacturer lasts between 30,000 miles and 50,000 miles

=P(30000

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

2) Given that the daily high temperature on October 31 in a certain city is normally distributed with 50 and 8

Y - daily high temp on Oct 31 is N(50,8)

Or Z = \frac{x-50}{8} is N(0,1)

the probability that a randomly chosen tire from this manufacturer lasts between 46 degrees and 58 degrees

=P(46

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$42.80

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The antibiotic clarithromycin is eliminated from the body according to the formula A(t) = 500e−0.1386t, where A is the amount re
PolarNik [594]

Answer:

Time(t) = 11.61 hours (Rounded to two decimal place)

Step-by-step explanation:

Given: The antibiotic  clarithromycin is eliminated from the body according to the formula:

A(t) = 500e^{-0.1386t}                 ......[1]

where;

A - Amount remaining in the body(in milligram)

t - time in hours after the drug reaches peak concentration.

Given: Amount of drug in the body is reduced to 100 milligrams.

then,

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100= 500e^{-0.1386t}

Divide both sides by 500 we get;

\frac{100}{500}=\frac{ 500e^{-0.1386t}}{500}

Simplify:

\frac{1}{5} = e^{-0.1386t}

Taking logarithm both sides with base e, then we have;

\log_e (\frac{1}{5})= \log_e (e^{-0.1386t})

\log_e (\frac{1}{5})=-0.1386t         [ Using \log_e e^a =a ]

or

\log_e (0.2)=-0.1386t

-1.6094379124341 = -0.1386t

 [using value of \log_e (0.2) = -1.6094379124341 ]

then;

t = \frac{-1.6094379124341}{-0.1386}

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t ≈11.61 hours.

Therefore, the time 11.61 hours(Rounded two decimal place) will pass before the amount of drug in the body is reduced to 100 milligrams


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1. Find each measure.<br> a)<br> m20<br> 8<br> R<br> Select<br> 110"<br> S<br> 8<br> T<br> c)
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Answer:

pls write question correctly so I can answer it

7 0
2 years ago
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The answer is 1;3 because if you simplified 2;6 you get 1;3 so the answer to that is
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