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nlexa [21]
2 years ago
10

There is a 70% chance that your car will get stuck in the snow during the next big snowfall. Given that you are already stuck in

the snow with your car, the chance that you will require a tow truck to pull you out is 90%. What is the chance that you will get stuck in the snow with your car and will require a tow truck to pull you out?
Hint: cap p left parenthesis cap a vertical line cap b right parenthesis equals start fraction cap p left parenthesis cap a intersection cap b right parenthesis over cap p left parenthesis cap b right parenthesis end fraction

29%
63%
78%
66%
Mathematics
1 answer:
Flura [38]2 years ago
6 0
Given:
Probability that your car will get stuck in the snow: 70%
Probability that you will require a tow truck when stuck in the snow is 90%.

Probability that you're stuck in the snow and need a tow truck.

70% x 90% = 63%  2nd option.

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Answer:

Mean = 15.15

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a = 14.1

b = 16.2

a) Mean:

\mu = \displaystyle\frac{a+b}{2}\\\\\mu = \frac{14.1+16.2}{2} = 15.15

b) Standard Deviation:

\sigma = \sqrt{\displaystyle\frac{(b-a)^2}{12}}\\\\= \sqrt{\dfrac{(16.2-14.1)^2}{12}} = \sqrt{0.3675} = 0.6062

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3 years ago
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,

Substitute the value of A = 100 milligrams in [1] we get;

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}

Simplify:

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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2 years ago
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