Let F(s) 1.1s + 0.03s2 represent the stopping distance (in feet) of a car travelling at s miles per hour. Calculate F(60) and es
1 answer:
Answer:
F(60) = 3666 feet
Estimated Increase in Stopping Distance: 123.1 feet
Actual Increase in Stopping Distance: 122.1 feet
The cost of producing 3000 bagels is $1036.5
The Estimated cost of the 3001st bagel is $1036.736.
The actual cost of the 3001 st bagel is $1036.737.
Step-by-step explanation:
F(s) = 1.1s + s²
F(60) = 1.1(60) + (60)²
= 66 + 3600
F(60) = 3666 feet
To find the estimate increase in stopping distance, differentiate the function to get F'(s) and then find F'(61)
F'(s) = = 1.1 + 2s
F'(61) = 1.1 + 2(61)
F'(61) = 123.1 feet
If speed is increased from 60 to 61, we can find the actual increase by finding F(61) and then subtracting F(60) from it.
F(61) = 1.1(61) + (61)²
F(61) = 3788.1 feet
Increase = 3788.1 - 3666
Increase = 122.1 feet
F(60) = 3666 feet
Estimated Increase in Stopping Distance: 123.1 feet
Actual Increase in Stopping Distance: 122.1 feet
C(x) = 300 + 0.25x - 0.5 ()³
C(3000) = 300 + 0.25(3000) - 0.5 (3000/1000)³
= 300 + 750 - 13.5
C(3000) = $1036.5
The cost of producing 3000 bagels is $1036.5
To estimate the cost of the 3001st bagel, we need to differentiate the function and then find the increase in price at the 3001st bagel. The answer then needs to be added to C(3000). So,
C'(x) = 0.25 - 0.5*3 (x/1000)²/1000
C'(3001) = 0.25 - 0.5*3 (3001/1000)²/1000
C'(3001) = 0.236
C(3001) = C(3000) + C'(3001)
= 1036.5 + 0.236
C(3001) = $1036.736
The Estimated cost of the 3001st bagel is $1036.736.
C(3001) = 300 + 0.25(3001) - 0.5(3001/1000)³
= 300 + 750.25 - 13.5135
C(3001) = $1036.737
The actual cost of the 3001 st bagel is $1036.737.
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a)
b)
c)
Sicne this probability just represent 1% of the data we can consider this value as unusual.
d)
And if we solve for a we got
So the value of height that separates the bottom 90% of data from the top 10% is 87360.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:
Where and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using excel or the normal standard table and we got:
Part b
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule and excel or the normal standard table and we got:
Part c
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule and excel or the normal standard table and we got:
Sicne this probability just represent 1% of the data we can consider this value as unusual.
Part d
For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 90% of data from the top 10% is 87360.