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qwelly [4]
2 years ago
9

Describe the function over each part of its domain. State whether it is constant, increasing, or decreasing, and state the slope

over each part.

Mathematics
2 answers:
ladessa [460]2 years ago
8 0
It would be increasing, because the numbers are getting higher.
SVETLANKA909090 [29]2 years ago
4 0

Answer:

 domain (0,12000) is constant and slope =0

domain (12000, 38000) is decreasing and slope = negative

Step-by-step explanation:

We analyze the given graph

From x=0  to x=12,000 the y values remains the same

the value of y is constant, there is no increase or decrease in y

So from x=0 to x= 12000, the graph is cost is constant

Hence slope =0

 domain (0,12000) is constant and slope =0

From x=12,000  to x=38,000 the y values is decreasing

For decreasing y values the slope is negative

So we can say the slope is negative

domain (12000, 38000) is decreasing and slope = negative

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Johann fits a linear model to a set of data. The line of best fit has the equation y=0.331986x+5.33286. The scatter plot with th
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Answer:

6.33

Step-by-step explanation:

Here, we are asked to give a residual value at x = 3

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The line of best fit equation according to the question is;

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Substituting x here, we have

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