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2 years ago

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Sam has been asked to prepare 40 documents in 2 work days(there are 8 hours in each workday).Each document takes 30 minutes to p

repare.Assuming Sam works both days,how many hours of help will he need to finish in 2 work days?

1 answer:

sladkih [1.3K]2 years ago

7 0

Answer:n 2 work days, there are 8 hours in each workday. How many hours of help will he need in 2 work days

Step-by-step explanation:

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2 years ago

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find the probability that a randomly selected automobile tire has a tread life between 42000 and 46000 miles

maria [59]

Given that in a national highway Traffic Safety Administration (NHTSA) report, data provided to the NHTSA by Goodyear stated that the mean tread life of a properly inflated automobile tires is 45,000 miles. Suppose that the current distribution of tread life of properly inflated automobile tires is normally distributed with mean of 45,000 miles and a standard deviation of 2360 miles.

Part A:

Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
$P(a \ \textless \ X \ \textless \ b) = P(X \ \textless \ b) - P(X \ \textless \ a) \\ \\ P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
$P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\ \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}$

b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
$P(X \ \textgreater \ a) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
$P(X \ \textgreater \ 50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\ \\ =1-0.98294=\bold{0.0171}$

Part C:

Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
$P(X \ \textless \ a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
$P(X \ \textless \ 38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\ \\ =P(z\ \textless \ -2.966)=\bold{0.0015}$

d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
$P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.

Thus:
$P(X \ \textgreater \ x) =0.06 \\ \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\ \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\ \\ \Rightarrow x=3,669.8+45,000=48,669.8$
Therefore, the value of x is 48,669.8

e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
$P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.

Thus:
$P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56$
Therefore, the value of x is 40,152.56

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2 years ago

Nina and Ryan each ran at a constant speed for a 100-meter race. Each runner’s distance fro the same section of the race is disp

sergey [27]

Step-by-step explanation:

Ryan had a head start of 10 meters.

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2 years ago

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