In this exercise we have to use the knowledge of algorithm to write the correct alternative that best matches, thus we can say that:
Letter D
The computational complicatedness of K-NN increases as the extent or bulk of some dimension of the training basic document file increase and the treasure gets considerably unhurried as the number of examples and free variables increase.
Also, K-NN happen a non-parametric machine intelligence treasure and as such form no assuming possession about the working form of the question at hand.
The invention everything better accompanying information in visible form of the same scale, therefore standard the information in visible form superior to applying the invention happen urged.
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All the statements provided about k-NN (scale importance, efficiency with small input variables, and no assumptions about the functional form of the problem) are true.
Explanation:The question asks which of the given statements is true about the k-NN (k-Nearest Neighbors) algorithm. The correct answer would be 'D) All of the above.' This is because all three statements accurately describe different aspects of how the k-NN algorithm works:
k-NN performs much better if all of the data have the same scale. Applying a suitable scaling method is important as it ensures all features contribute equally. k-NN works well with a small number of input variables (p), but struggles when the number of inputs grows large. This is due to the 'curse of dimensionality,' as the volume of the input space increases exponentially with the number of dimensions, impacting performance and accuracy. k-NN makes no assumptions about the functional form of the problem being solved. This is a key strength of k-NN; it's a non-parametric method that doesn’t make explicit assumptions about the functional form of the data.Learn more about k-NN algorithm here:
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A wire 6 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures?
Give the length of wire used for each:
For the square:
For the circle:
(for both, include units)
Where should the wire be cut to maximize the total area? Again, give the length of wire used for each:For the square:For the circle:(for both, include units)
Answer:
Used wire in circle x = 2.64 m
Used in square L - x = 3.36 m
Total wire used 6 m
Step-by-step explanation:
We have a wire of 6 meters long.
We will cut it a distance x from one end, to get two pieces
x and 6 - x
We are going to use the piece x to get the circle then
So Perimetr of a circle is 2π*r (r is the radius of the circle) then:
x = 2*π*r ⇒ r = x/2*π
And area would be A(c) = π* (x/2*π)² ⇒ A(c) = x²/4π
From 6 - x we will get a square, and as the perimeter is 4 times the side
we have
( 6 - x )/ 4 is the side of the square
And the area is A(s) = [( 6 - x ) /4]²
Total area as function of x is
A(t) = A(c) + A(s)
A(x) = x²/4π + [ ( 6 - x ) / 4 ]²
A(x) = x²/4π + (36 + x² - 12x) /16
A(x) = 1 / 16π [ 4x² + 36π + πx² - 12π x ]
Taking drivatives on both sides of the equation we get:
A´(x) = 1/ 16π [8x +2πx - 12π]
A´(x) = 0 ⇒ 1/ 16π [8x +2πx - 12π] = 0
[8x +2πx - 12π] = 0
8x + 6.28x - 37.68 = 0
14.28x - 37.68 = 0 ⇒ x = 37.68 /14.28
x = 2.64 m length of wire used in the circle
Then the length L for the side of the square is
(6 - x )/4 ⇒ ( 6 - 2.64 )/ 4 ⇒ 3.36 / 4
L = 0.84 m total length of wire used in the square is
3.36 m
And total length of wire used is 6 m
The function is a quadratic function and "a" coefficient is positive then is open upward parabola there is not a maximun
Answer:
Wire used in circle , x = 2.64 m
Wire used in square, L - x = 3.36 m
Total used wire is 6 m
Step-by-step explanation:
We have a wire of 6 metres long.
We will cut it a distance x metre from one end, to get two pieces x metre and 6 - x metres.
We are going to use the piece of x metre to get the circle
So, Perimeter of the circle is [tex]2\pi r[/tex] (r is the radius of the circle) then
[tex]x = 2\pi r[/tex] ⇒ [tex]r = \frac{x}{2} \pi[/tex]
And area would be [tex]A(c) =\pi (\frac{x}{2} \pi )^{2}[/tex]⇒[tex]A(c) = \frac{x^{2} }{4\pi }[/tex]
From [tex]6 - x[/tex] we will get the square, and as the perimeter is 4 times the side
we have
[tex]\frac{6 - x}{4}[/tex] is the side of the square
and the area is [tex]A(s) = (\frac{6 - x}{4}) ^{2}[/tex]
Total area of the function of x is
[tex]A(t) = A(c) + A(s)[/tex]
[tex]A(x) = \frac{x^{2} }{4\pi } +(\frac{6 - x}{4} )^{2}\\A(x) = \frac{x^{2} }{4\pi } + \frac{36+x^{2} -12x}{16} \\A(x) = \frac{1}{16\pi } (4x^{2} +36\pi +\pi x^{2} -12\pi x)[/tex]
Taking derivative on the both side of the equation we get :
[tex]A^{'} = \frac{1}{16\pi } (8x+2\pi x-12\pi )\\[/tex]
[tex]A^{'} = 0[/tex]
[tex]\frac{1}{16\pi } (8x+2\pi x-12\pi ) = 0\\(8x+2\pi x-12\pi ) = 0\\8x + 6.28x-37.68=0\\14.28x-37.68=0\\x=2.64 m[/tex]
length of wire used in the circle is x = 2.64 m
Then the length L of the wire used in the square is
[tex]\frac{6 - x}{4}[/tex] ⇒[tex]\frac{6 - 2.64}{4}[/tex] ⇒ [tex]\frac{3.36}{4}[/tex]
L = 0.84 m
Total length of the wire used in the square is 4L = 3.36 m
And total length of the wire used is 6 m
The function is a quadratic function and "a" coefficient is positive then is open upward parabola there is not a maximum.
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If the recommended adult dosage for a drug is D (in mg), then to determine the appropriate dosage c for a child of age a, pharmacists use the equation c = 0.0417D(a + 1). Suppose the dosage for an adult is 200 mg.
(a) Find the slope of the graph of c. (Round your answer to two decimal places.)
(b) What is the dosage for a newborn? (Round your answer to two decimal places.)
Answer:
a) The slope of the graph of c is 8.34.
b) The dosage for a newborn is 8.34mg.
Step-by-step explanation:
A first order function in the following format
[tex]c(a) = ba + d[/tex]
Has slope b.
The appropriate dosage c is a function of the age a.
In this problem, we have that:
[tex]c(a) = 0.0417D(a + 1)[/tex]
Suppose the dosage for an adult is 200 mg. This means that [tex]D = 200[/tex]
So
[tex]c(a) = 0.0417*200(a + 1)[/tex]
[tex]c(a) = 8.34a + 8.34[/tex]
(a) Find the slope of the graph of c.
The slope of the graph of c is 8.34.
(b) What is the dosage for a newborn? (Round your answer to two decimal places.)
A newborn has age a = 0. So this is c(0).
[tex]c(a) = 8.34a + 8.34[/tex]
[tex]c(0) = 8.34*0 + 8.34 = 8.34[/tex]
The dosage for a newborn is 8.34mg.
Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15 (this should be used as the baseline for the questions below.) (a) If one person is randomly selected from the general population, what is the probability of getting someone with an IQ of at least 133?
Answer:
a) [tex]P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z\geq 2.2)=1- P(z<2.2)=1-0.986=0.0139 [/tex]
b) [tex]P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z\geq 6.6)=1- P(z<6.6)=1-0.999999=2.06x10^{-11} [/tex]
c) No. The mean can be lower than 131.5 if we find the probability:
[tex]P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)[/tex]
[tex]P(Z\leq 6.3) \approx 1 [/tex]
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".
A. If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,15)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]
We are interested on this probability
[tex]P(X\geq 133)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z\geq 2.2)=1- P(z<2.2)=1-0.986=0.0139 [/tex]
And the probability is calculated from the normal standard table or with excel.
B. If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.
For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:
[tex] \bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)[/tex]
The new z score is defined as
[tex]z=\frac{x-\mu}{\sigma_{\bar x}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z\geq 6.6)=1- P(z<6.6)=1-0.999999=2.06x10^{-11} [/tex]
And the probability is calculated from the normal standard table or with excel.
C. Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?
No. The mean can be lower than 131.5 if we find the probability:
[tex]P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)[/tex]
[tex]P(Z\leq 6.3) \approx 1 [/tex]
Answer:
0.0139
Step-by-step explanation:
We have to find P(X≥133)
IQ scores normally distributed with mean=100 and standard deviation=sd=15
[tex]P(X\geq 133)=P(\frac{x-mean}{sd} \geq \frac{133-100}{15}[/tex]
[tex]P(Z\geq 2.2)=P(0<Z<infinity)-P(0<z<2.2)[/tex]
[tex]P(Z\geq 2.2)=0.5-0.4861[/tex]
[tex]P(Z\geq 2.2)=0.0139[/tex]
P(X≥133)=0.0139.
So, if one person is selected at random the probability of getting someone with at least 133 IQ score is 1.39%
If cos theta = -2/3 which of the following are possible?
A. sin theta = - (sq5)/3 and tan theta = (sq5)/2
B. csc theta = 3/(sq5) and tan theta = -(sq5)/2
C. csc theta = -3/2 and tan theta = (sq5)/2
D. sin theta = (sq5)/3 and tan theta = (sq5)/2
Answer:
sin theta = opp/hyp = √5/3 = (sq5)/3
csc theta = 1 / sin theta = 1 / (√5/3) = 1 x 3/√5 = 3/√5 = 3/(sq5)
csc theta = 3/(sq5) and tan theta = -(sq5)/2 = B
Step-by-step explanation:
if cos theta = -2/3
the -2 is adjacent
3 is the hypothenus
using pythagoras theorem
(hypothenus)² = (opposite)² + (adjacent)²
3² = opp²+ (-2)² = opp²+ (4)
9-4 = opp² = 5
opp = √5
sin theta = opp/hyp = √5/3 = (sq5)/3
tan theta = opp/adj = √5/-2 = (sq5)/2
csc theta = 1 / sin theta = 1 / (√5/3) = 1 x 3/√5 = 3/√5 = 3/(sq5)
csc theta = 3/(sq5) and tan theta = -(sq5)/2 = b
Answer:
A and B are correct!
Step-by-step explanation:
Opposite side= - or +(sq5)
Adjacent side= -2
Hypotenuse= 3
Drug x, a drug that claims to treat male pattern scalp hair loss, was administered for 12 months to over 1800 men aged 18 to 41 with mild to moderate amounts of ongoing hair loss. Whether they were receiving drug x or a placebo ( a pill containing no medication), all men were given a medicated shampoo. In general, men who took drug x maintained or increased the number of visible scalp hairs; while scalp hairs counts in men who took the placebo continued to decrease. This concluded that drug x is effective in maintaining or increasing the amount of scalp hair in men.a. Which statement in this example can be referred to as descriptive statistics?b. Which statement in this example can be referred to as inferential statistics?
The following statement
'In general, men who took drug x maintained or increased the number of visible scalp hairs; while scalp hairs counts in men who took the placebo continued to decrease'
can be referred as descriptive statistics, since we are saying that the msot of the men who took the drug gain scalp hairs while the contrary happened from those men who took placebo.
On the other hand, the statement
'drug x is effective in maintaining or increasing the amount of scalp hair in men'
Is a deduction you make, so it can be referred as inferential statistics
Find the width, in inches, of a 65 mm film. (Enter a number as an integer or decimal.)
Answer:
The answer to your question is 2.6 in
Step-by-step explanation:
Data
width = ? in
width = 65 mm
Process
1.- To solve this problem, use proportions and cross multiplication
1 in ---------------------- 25.4 mm
x ---------------------- 65 mm
x = (65 x 1) / 25.4
2.- Simplification
x = 65 / 25.4
3.- Result
x = 2.6 in
What is the correct negation of "There is someone in Spain who speaks at least 10 languages"?1) "Everyone in Spain speaks at least 10 languages."2) "Everyone outside of Spain speaks at least 10 languages."3) "There is someone in Spain who speaks at most 9 languages."4) "Everyone in Spain speaks at most 9 languages."5) "Everyone outside of Spain speaks at most 9 languages."6) "There is someone outside of Spain who speaks at most 9 languages."7) "There is someone outside of Spain who speaks at least 10 languages."
Answer:
4) "Everyone in Spain speaks at most 9 languages."
Step-by-step explanation:
A negation is a statement that contradicts the original statement. Since the original statement regards people in Spain, options 2, 5, 6 and 7 can be ruled out as they refer to speakers outside of Spain.
Evaluating the remaining statements.
1) "Everyone in Spain speaks at least 10 languages."
If that is true, then someone in Spain speaks at least 10 languages, it doesn't negate the original statement.
3) "There is someone in Spain who speaks at most 9 languages."
It only refers to one specific individual, it doesn't negate the original statement.
4) "Everyone in Spain speaks at most 9 languages."
This means that nobody in Spain speaks more than 9 languages, contradicting the original statement.
The answer is statement 4.
In a class of 125 students, 27 are computer science majors, 51 are mechanical engineering majors, 12 are civil engineers and the rest are general engineering majors. Assume students only have one major.
Suppose six students from the class are chosen at random what is the probability none are mechanical engineering majors?
I tried 74/125 * 73/124 * 72/123 * 71/122 * 70/121 * 69/120 * 68/119 but had incorrect answer
Answer:
There is a 3.95% probability that none are mechanical engineering majors.
Step-by-step explanation:
[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, we have that:
There are 125 students.
51 of them are mechanical engineering majors and 125-51 = 74 are not mechanical engineering majors.
Suppose six students from the class are chosen at random what is the probability none are mechanical engineering majors?
The total number of students is 125. So total number of 6 student groups is
[tex]C_{125,6} = \frac{125!}{6!119!}[/tex]
The total number of non mechanical engineering students is 74. So the total number of 6 non mechanical engineering students is.
[tex]C_{74,6} = \frac{74!}{6!68!}[/tex]
The probability is:
[tex]\frac{C_{74,6}}{C_{125,6}} = 74/125 * 73/124 * 72/123 * 71/122 * 70/121 * 69/120 = 0.0395[/tex]
The 68/119 ends up simplified in this exercise, this is your mistake.
There is a 3.95% probability that none are mechanical engineering majors.
Suppose Q is the midpoint of line segment PR, PQ = x + 10, and QR = 4x - 2
What is the value of PR?
A. 4
B. 8
C. 14
D. 28
answer and step by step in the attachment
The value of PR is 28.
PQ = x + 10
QR = 4x - 2
To solve the question, we'll equate both equations which will be:
PQ = QR.
x + 10 = 4x - 2
Collect like terms
4x - x = 10 + 2.
3x = 12
x = 12/3
x = 4
Therefore,
PQ = x + 10 = 4 + 10 = 14
QR = 4x - 2 = 4(4) - 2 = 16 - 2 = 14
Therefore, PR = 14 + 14 = 28
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Math problem is on the attached picture. Thanks!
Answer:
the consecutive integars is 11,12,13,14,15
Step-by-step explanation:
let the first integar be described as y since they are consecutive ( following each other)the second will y + 1the third is y + 2the fourth is y + 3the fifth y + 4then it says the sum of the first and 4 times the third = 60 less than 3 times the sum of the second, fourth and fifth
60 less than 3 times the sum of the second, fourth and fifth means that 3 times the sum of the second, fourth and fifth minus 60
y+ 4(y+2) = 3{ (y +1) +(y+3) + (y+4)} -60
open the brackety + 4y + 8 =3[ y +1 + y + 3 +y+4} - 60
5y +8=3(y+y+y+1+3+4)-60
5y+8=3(3y+8)-60
simplify and open the brackets5y+8=9y+24-60
5y+8=9y-36
subtract 8 from both sides5y+8-8=9y-36-8
5y=9y-44
subtact 9y from both sides5y-9y=9y-9y-44
-4y=-44
divide -4 from both sides-4y/-4=-44/-4 ( minus divided by minus is plus
y = 11
the first integar is y = 11the second will y + 1 = 11+1 = 12the third is y + 2 = 11 +2 = 13the fourth is y + 3 = 11 + 3 = 14the fifth y + 4 = 11 + 4 = 15the consecutive integars is 11,12,13,14,15How many possible birth orders with respect to gender are possible in a family with six children
Answer:
There are 64 possible birth orders in a family with six children.
Step-by-step explanation:
Let's start from 1 children.
You can have M or F. So two possible birth orders.
With two children, you can have M-M, M-F, F-M, F-F. So 2^2 = 4 possible birth orders.
For three children.
M-M-M, M-M-F, M-F-M, M-F-F, F-M-M, F-M-F, F-F-M, F-F-F. So 2^3 = 8 possible birth orders.
Generalizing:
For n children, you can have [tex]2^{n}[/tex] birth orders.
In this problem, we have that:
The family has 6 chilren.
So there are [tex]2^{6} = 64[/tex] possible birth orders.
Determine if the statement is true or false:
1. If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.
2.Different sequences of row operations can lead to different echelon forms for the same matrix.
3.Different sequences of row operations can lead to different reduced echelon forms for the same matrix.
4.If a linear system has four equations and seven variables, then it must have infinitely many solutions.
Answer:
Step-by-step explanation:
given are four statements and we have to find whether true or false.
.1 If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.
True
2.Different sequences of row operations can lead to different echelon forms for the same matrix.
True in whatever way we do the reduced form would be equivalent matrices
3.Different sequences of row operations can lead to different reduced echelon forms for the same matrix.
False the resulting matrices would be equivalent.
4.If a linear system has four equations and seven variables, then it must have infinitely many solutions.
True, because variables are more than equations. So parametric solutions infinite only is possible
Statements 1 and 2 are true, and statements 3 and 4 are false. While equivalent matrices can be transformed into each other and different row operations can yield different echelon forms, the reduced echelon form is unique, and a linear system with more variables than equations does not necessarily have infinitely many solutions.
Explanation:1. The statement is true. If two matrices are equivalent, one can indeed be transformed into the other through a sequence of elementary row operations. These operations include swapping two rows, multiplying a row by a non-zero scalar, or adding a multiple of one row to another row.
2. The statement is true. Different sequences of row operations can yield different echelon forms of the same matrix as the operations can redistribute information about the system of equations in different ways.
3. The statement is false. Regardless of the sequence of row operations performed, the reduced echelon form of a matrix is unique. This is because the reduced echelon form is a canonical form, meaning there's only one possible reduced echelon form for a given matrix.
4. The statement is false. Even though a linear system has more variables than equations, this does not guarantee that it will have infinitely many solutions. It could have no solutions or, under certain conditions, even a unique solution. Further analysis is required in these cases.
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A supermarket has determined that daily demand for milk containers has an approximate bell shaped distribution, with a mean of 55 containers and a standard deviation of six containers. How often can we expect between 49 and 61 containers to be sold in a day?
Answer:
[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.
Step-by-step explanation:
The daily demand for milk containers has a Normal (or Gaussian) distribution, and we can use values from the cumulative distribution function and z-scores to solve the question.
We know from the question that the mean of the distribution is:
[tex] \\ \mu = 55 [/tex]
And a standard deviation of:
[tex] \\ \sigma = 6 [/tex]
The z-scores permit calculates the probabilities for any case whose values have a Normal o Gaussian distribution. Then, for this, we need to calculate the z-scores for 49 containers and 61 containers to establish the corresponding probabilities, as well as the differences between these two values to determine the probability between them.
These z-scores are given by:
[tex] \\ z = \frac{x-\mu}{\sigma} [/tex]
Thus,
The z-scores for 49 and 61 containers are:
[tex] \\ z_{49} = \frac{49 - 55}{6} = \frac{-6}{6} = -1 [/tex] [1]
[tex] \\ z_{61} = \frac{61 - 55}{6} = \frac{6}{6} = 1 [/tex] [2]
Well, this is a special case when in both cases the values are one standard deviation from the mean, but in one case ([tex] \\ z_{49} = -1 [/tex]) the values are smaller than the mean and in the other case ([tex] \\ z_{61} = 1 [/tex]) the values are greater than the mean.
In other words, the cumulative probability for ([tex] \\ z_{61} = 1 [/tex]), obtained from any Table of the Normal Distribution available on the Web, is: 0.8413 (or 84.13%) and the cumulative probability for ([tex] \\ z_{49} = -1 [/tex]) is: 1 - 0.8413 = 0.1587 (or 15.87%), because of the symmetry of the Normal Distribution.
Then, the probability of expecting to sell between 49 and 61 containers in a day is the difference of both obtained probabilities:
[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.
See the graph below.
Final answer:
Approximately 68% of the time, the number of milk containers sold in a day at the supermarket is expected to be between 49 and 61 containers, following the Empirical Rule for normal distribution.
Explanation:
The question asks us about the probability of daily sales being between certain values when they follow a bell-shaped distribution, specifically between 49 and 61 containers. Since we know the distribution is approximately normal with a mean of 55 containers and a standard deviation of six containers, we can use the Empirical Rule or the Standard Normal Distribution to find the probability.
In the context of a normal distribution, we know that:
About 68% of the data falls within one standard deviation of the mean. (Mean ± 1SD)
About 95% falls within two standard deviations. (Mean ± 2SD)
For this supermarket, one standard deviation from the mean (55 ± 6) gives us a range from 49 to 61 containers. Therefore, approximately 68% of the time, we can expect the number of milk containers sold to be between 49 and 61 containers in a day, based on the Empirical Rule.
Determine whether the statement is true or false. You toss a coin and roll a die. The event "tossing tails and rolling a 4 or 6" is a simple event.
Answer: False.
Step-by-step explanation:
The probability for any event = [tex]\dfrac{\text{favorable outcomes}}{\text{Total outcomes}}[/tex]
Definition : A simple event is an event that has only one possible favorable outcomes.
If we toss a coin and roll a die .
The possible outcomes for event "tossing tails and rolling a 4 or 6" = {(T, 4) , (T,6)}
Since this event has 2 possible favorable outcomes , Therefore it is not an simple event.
Hence, the given statement is "false".
The statement 'tossing tails and rolling a 4 or 6' is not a simple event. A simple event only consists of a single outcome, whereas this case includes multiple outcomes: getting tails in the coin toss and either a 4 or 6 in the die roll. As such, the statement is false.
Explanation:The statement 'tossing tails and rolling a 4 or 6' is false. This is because it is not a simple event. In probability theory, a simple event is an event that consists of a single outcome. However, in this case, there are multiple outcomes - getting a tails on the coin toss, and getting either a 4 or 6 on the die roll.
For example, the possible sets are: {Tails, 4} and {Tails, 6}. This indicates that there are multiple outcomes within this event, thus it cannot be considered a simple event.
Hence, the given statement is false because a single outcome in this scenario could only be {Tails, 4} or {Tails, 6}, not both of them combined.
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A manufacturer of pharmaceutical products analyzes specimens from each batch of a product to verify the concentration of the active ingredient, required to be .80 grams per liter. Suppose the repeated measurements follow a Normal distribution. Fifteen specimens are taken from a batch and it is found that 0.82 grams per liter and s 0.048 grams per liter. Find a 99% confidence interval for μ, the true average concentration of the active ingredient.
Answer:
[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]
[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]
So on this case the 99% confidence interval would be given by (0.783;0.857)
So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
[tex]\bar X= 0.82[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=0.048 represent the sample standard deviation
n=15 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=15-1=14[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that [tex]t_{\alpha/2}=2.977[/tex]
Now we have everything in order to replace into formula (1):
[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]
[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]
So on this case the 99% confidence interval would be given by (0.783;0.857)
So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval
To find a 99% confidence interval for the true average concentration of the active ingredient, we can use the formula: CI = (x - z*(s/sqrt(n)), x + z*(s/sqrt(n))). Plugging in the given values and using the critical value for a 99% confidence level, we get the confidence interval to be (0.799, 0.841) grams per liter.
Explanation:To find a 99% confidence interval for the true average concentration of the active ingredient, we can use the formula:
CI = (x - z*(s/sqrt(n)), x + z*(s/sqrt(n)))
Where CI is the confidence interval, x is the sample mean, z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.
Plugging in the given values: x = 0.82, s = 0.048, and n = 15, and using the critical value for a 99% confidence level (z = 2.576), we get:
CI = (0.82 - 2.576*(0.048/sqrt(15)), 0.82 + 2.576*(0.048/sqrt(15)))
Simplifying the expression, we get the confidence interval to be (0.799, 0.841) grams per liter.
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Suppose that the average and standard deviation of the fine for speeding on a particular highway are 111.12 and 13.04, respectively. Calculate an interval that is symmetric around the mean such that it contains approximately 68% of fines. Assume that the fine amount has a normal distribution. 1) (124.16, 98.08) 2) (111.12, 13.04) 3) (98.08, 124.16) 4) (85.04, 137.2) 5) (137.2, 85.04)
Answer:
3) (98.08, 124.16)
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 111.12
Standard deviation = 13.04
Calculate an interval that is symmetric around the mean such that it contains approximately 68% of fines.
68% of the fines are within 1 standard deviation of the mean speed. So
From 111.12 - 13.04 = 98.08 to 111.12 + 13.04 = 124.16
The interval notation in the smallest value before the highest value.
So the correct answer is:
3) (98.08, 124.16)
As part of the National Health and Nutrition Examination Survey, subjects were asked how long they slept the preceding night, and the following times (hours) were reported: 8, 7, 5, 7, 4, 7, 6, 7, 8, 8, 8, 6
The standard deviation is 1.3 hours.
What is the variance?
Final answer:
The variance for the given data is 1.69 hours squared.
Explanation:
To find the variance, we need to square the standard deviation. The standard deviation is given as 1.3 hours, so when we square it, we get 1.69 hours squared. Therefore, the variance for the given data is 1.69 hours squared.
Billy is walking from the front door of his house to his bus stop, which is 960 feet away from his front door. As Billy walks out his front door, he walks in a straight path toward his bus stop at a constant rate of 7.5 feet per second.
(A) Illustrate the situation with a diagram and define variables to represent the values of the relevant varying quantitities. (Label the variables on your picture.)
(B) Define a function f to determine Billy's distance from his bus stop in terms of the number of seconds he has been walking.
(C) What is the independent quantity and what is the domain of f (the values the independent quantity can take on)?
(D) What is the dependent quantity and what is the range of f (the values the dependent quantity can take on)?
(E) What do each of the following represent: f(0) and f(60.25)?
Answer:
b) 690 - 7.5*t
c) 0 < t < 92s time (t) is independent quantity
d) 0 < s < 690ft distance from bus stop (s) is dependent quantity
e) f(0) = 690 ft away from bus stop , f(60.25) = 238.125 ft away from bus stop
Step-by-step explanation:
Part a - see diagram
part b
initial distance from bus stop s0 = 690 ft
distance covered = 7.5*t
s = s0 - distance covered
s = 690 - 7.5*t = f(t)
part c
s = 0 or s = 690
0 = 690 -7.5*t
t = 92 s
Hence domain : 0 < t < 92s time (t) is independent quantity
part d
s = 0 or s = 690
Hence range : 0 < s < 690ft distance from bus stop (s) is dependent quantity because it depends on time (t)
part e
f(0) is s @t = 0
f(0) = 690 ft away from bus stop
f(60.25) is s @t = 60.25
f(60.25) = 690 - 7.5*60.25 = 238.125 ft away from bus stop.
What is the additive inverse of the complex number -8 + 3i?
-8-3i
-8+3i
8-3i
8+3i
Answer:
cc
Step-by-step explanation:
The additive inverse of the complex number -8 + 3i is 8 - 3i.
What are complex Numbers?The complex number is basically the combination of a real number and an imaginary number.
The additive inverse of a complex number is the number that, when added to the original number, results in zero.
Or it is the negative of the original number.
To find the additive inverse of the complex number -8 + 3i
we need to negate both the real and imaginary parts. So:
Additive inverse = -( -8 + 3i)
= 8 - 3i
Therefore, the additive inverse of the complex number -8 + 3i is 8 - 3i.
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how do I solve these three
Answer:
below
Step-by-step explanation:
2(x+5)=3x+1
2x+10=3x+1
10-1=3x-2x
9=x
x=9
3y-4=6-2y
3y+2y=6+4
5y=10
y=10/5
y=2
3(n+2)= 9(6-n)
3n+6=54-9n
3n+9n=54-6
12n=48
n=48/12
n=4
Answer:
Step-by-step explanation:
1) 2(x + 5) = 3x + 1
Multiplying each term inside the parenthesis by 2, it becomes
2x + 10 = 3x + 1
Subtracting 2x and - 1 from both sides of the equation
2x - 2x + 10 - 1 = 3x - 2x + 1 - 1
3x - 2x = 10 - 1
x = 9
2) 3y - 4 = 6 - 2y
Adding 4 and 2y to the left hand side and the right hand side of the equation, it becomes
3y + 2y - 4 + 4 = 6 + 4 + 2y - 2y
5y = 10
Dividing the both sides of the equation by 5, it becomes
5y/5 = 10/5
y = 10/2 = 5
3) 3(n + 2) = 9(6 - n)
3n + 6 = 54 - 9n
Subtracting 6 from both sides and adding 9n to both sides of the equation, it becomes
3n + 9n + 6 - 6 = 54 - 6 - 9n + 9n
12n = 48
Dividing both sides of the equation by 12, it becomes
12n/12 = 48/12
n = 4
A grocery store manager must decide how to best present a limited supply of milk and cookies to its customers. Milk can be sold by itself for a profit of $1.50 per gallon. Cookies can likewise be sold at a profit of $2.50 per dozen. To increase appeal to customers, one gallon of milk and a dozen cookies can be packaged together and are then sold for a profit of $3.00 per bundle. The manager has 100 gallons of milk and 150 dozen cookies available each day. The manager has decided to stock at least 75 gallons of milk per day and demand for cookies is always 140 dozen per day. To maximize profits, how much of each product should the manager stock.Which of the following is the objective function for the grocer's problem?A) Max P = 1.5M + 2.5C + 3BB) Min P = 1.5M + 2.5C + 3BC) Max P = 2.5M + 1.5C + 3BD) Max P = 2.5M + 3C + 1.5BE) Min P = 1.5M + 1.5C + 3B
Answer:
Option A.
Step-by-step explanation:
Let M be the milk per gallon.
C be the cookies per dozen.
B be the bundle (one gallon of milk and a dozen cookies ).
Milk can be sold by itself for a profit of $1.50 per gallon. Cookies can likewise be sold at a profit of $2.50 per dozen and bundle is sold for a profit of $3.00 per bundle.
[tex]Profit= 1.5M + 2.5C + 3B[/tex]
We need to maximize the profits. So, our objective function is
Therefore, the correct option is A.
Your city is represented in a coordinate plane where each unit represents 1 kilometer. The library is at (−2, −2), the post office is at (2, 2), and your house is at (−3, 2). You ride your bike from your house to the library, then the post office, and then back home. What is the minimum distance that you can ride your bike? Round your answer to the nearest tenth.
By using the distance formula, we calculate the distances between the house, library, and post office, finding the total minimum distance to be approximately 14.8 kilometers.
Explanation:This question requires the use of the distance formula in mathematics, which is derived from the Pythagorean theorem. The formula is √[(x₂ - x₁)² + (y₂ - y₁)²]. To find the total minimum distance you ride your bike, you calculate the distance from your house to the library, then the library to the post office, and finally, the post office back to your house.
First, calculate the distance from your house at (-3, 2) to the library at (-2, -2): √[(-2 - (-3))² + (-2 - 2)²] = √[1² + -4²] = √[1 + 16] = √17 ≈ 4.1 kilometers.
Second, calculate the distance from the library at (-2, -2) to the post office at (2, 2): √[(2 - (-2))² + (2 - (-2))²] = √[4² + 4²] = √[16 + 16] = √32 ≈ 5.7 kilometers.
Finally, calculate the distance from the post office at (2, 2) back to your house at (-3, 2): √[(-3 - 2)² + (2 - 2)²] = √[-5² + 0²] = √[25 + 0] = √25 = 5 kilometers.
Adding these distances together, 4.1 km + 5.7 km + 5 km = 14.8 kilometers.
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Two cards are drawn one at a time from a full deck of cards. a) What’s the probability they are an ace or a heart? b) What’s the probability they are neither ace nor a heart?
The probability of drawing an ace or a heart from a full deck of cards is approximately 5.88%. The probability of drawing a card that is neither an ace nor a heart is approximately 45.89%.
Explanation:This question is about probability in Mathematics. A full deck has 52 cards, 4 of which are aces, and 13 of which are hearts. The total number of favorable outcomes are the scenarios where we draw an ace or a heart.
For part a) There are 16 favorable outcomes (4 aces + 12 remaining hearts). For 2 draws, that's 16/52 * 15/51 = 0.0588 or 5.88% chance of drawing an ace or a heart.
For part b) There are 36 non-favorable outcomes (52 total - 16 favorable). The chance of drawing a card that is neither ace nor heart is 36/52 * 35/51 = 0.4589 or 45.89%.
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Suppose the reaction temperature X (in deg C) in a certainchemical process has uniform distribution with A= -5 and B=5.
a. Compute P(X<0)
b. Compute P(-2.5
c. Compute P(-2
Answer:
Step-by-step explanation:
Given that the reaction temperature X (in deg C) in a certainchemical process has uniform distribution with A= -5 and B=5
Thus the pdf of X would be
[tex]f(x) = 0.1 , -5\leq x\leq 5[/tex]
the cumulative probability
[tex]F(x) = \frac{x-a}{10} ,-5\leq x\leq 5[/tex]
a) P(X<0) = F(0) = [tex]\frac{0-(-5)}{10} =0.5[/tex]
b) P(x<-2.5) = F(-2.5)
= [tex]\frac{0-(-2,5)}{10} =0.25[/tex]
c) P(X>-2) = 1-F(-2)
=1-0.2 = 0.8
The volume of a cube is 7.14 cubic meters. What is the cross-sectional area that is parallel to one of its faces?
A = ________
Answer:
A = 3.706 square meters
Step-by-step explanation:
the length of a cube has equal side.
therefore, the volume of a cube is given by S³
V = S³ = 7.14
where S is the length of a side
the surface area of a cube is = 6S²
where the area a aside is calculated, then it is multiply by 6.
if S³ = 7.14
S = ∛(7.14) = 1.925 m
What is the cross-sectional area that is parallel to one of its faces?
this is saying we should calculate for the cross-sectional area of one face. All faces in a cube is equal and parallel to each other.
the crossectional area of one side of a cube is = S² = 1.925² = 3.706 square meters
A = 3.706 square meters
3. when the amount paid for land is $50,000 and the amount paid for expenses is $18,000, the balance in total assets after transaction (b) is $
Answer:
68,000
Step-by-step explanation:
$50,000 + $18,000 = $68,000
and the level of measurement for the educationeducation of the respondents in a survey classified as Institutional comma Autodidactic comma or OtherInstitutional, Autodidactic, or Other. Explain your choice.
The Question is understood as asking for [the Level of] Education of the respondents in a survey classified as [with values] Institutional, Autodidactic, or Other.
Answer:
The level of measurement is the Nominal Scale.
Step-by-step explanation:
The variable studied here is Education (of the respondents), and it is measured using three categories: Institutional, Autodidactic or Other.
As can be seen, the variable Education is measured using those categories that act as labels. These labels are simply names and they have neither relation in order to the other categories ---that is, no value is higher or lower than other--- and nor numerical meaning at all, as it is with other levels of measurement like Ordinal, Interval or Ratio.
For instance, having an Institutional value is not saying that it represents a higher value than having Autodidactic value or Other value. Moreover, if we substituted Institutional, Audidactic and Other by 0, 1, or 2, these values have no numerical value but are a way to classify the different possible values for the Education variable.
In other words, they represent only a way to classify values for the Education variable. No more than this.
As a result, the level of measurement for the variable Education is the Nominal Scale.
Construct the cumulative frequency distribution for the given data. Daily Low (degrees°F) Frequency 35-39 11 40-44 22 45-49 44 50-54 13 55-59 66 60-64 88 65-69 11 Construct the cumulative frequency distribution.
The cumulative frequency distribution of a data set is found by adding the frequency of each category to the sum of the frequencies of all previous categories. Using this method, the cumulative frequency distribution for the given data set is as follows: 35-39: 11, 40-44: 33, 45-49: 77, 50-54: 90, 55-59: 156, 60-64: 244, 65-69: 255.
Explanation:To construct the cumulative frequency distribution for the given data, we need to take into account the sum of all frequencies to the current one in addition to its own frequency. Let's look at how this works using the provided data:
35-39: 11 (there are no previous categories, so the cumulative frequency is 11)40-44: 22 + 11 = 3345-49: 44 + 33 = 7750-54: 13 + 77 = 9055-59: 66 + 90 = 15660-64: 88 + 156 = 24465-69: 11 + 244 = 255Therefore, the cumulative frequency distribution for the given data set is:
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A cumulative frequency distribution adds each frequency to the cumulative total of the previous ones. The sequence of cumulative frequencies for the given temperature data is 11, 33, 77, 90, 156, 244, and 255 respectively.
To construct a cumulative frequency distribution, we add up the frequencies as we go down the list of daily low temperatures. We will begin with the first frequency and add each subsequent frequency to the cumulative total of the previous frequencies. Here is how it would look for the data provided:
35-39: 11 (11)
40-44: 22 (11 + 22 = 33)
45-49: 44 (33 + 44 = 77)
50-54: 13 (77 + 13 = 90)
55-59: 66 (90 + 66 = 156)
60-64: 88 (156 + 88 = 244)
65-69: 11 (244 + 11 = 255)
The numbers in brackets represent the cumulative frequencies. Remember, to calculate Heating Degree Days (HDD) and Cooling Degree Days (CDD), you sum the number of days when the average temperature is below or above 65°F, respectively, multiplied by the difference from 65°F.
With its radar, an aircraft spots another aircraft 8000 feet away at a 12 degree angle of depression Determine the vertical distance and horizontal distance between the two aircraft.
Answer:the vertical distance is 1663.2 feet.
The horizontal distance is 7824.8 feet.
Step-by-step explanation:.
The vertical distance between the two aircrafts is represented by x.
To determine x, we would apply trigonometric ratio
Sin θ = opposite side/hypotenuse
Sine 12 = x/8000
x = 8000Sin12 = 8000 × 0.2079
x = 1663.2 feet
The horizontal distance between the two aircrafts is represented by y.
To determine y, we would apply trigonometric ratio
Cos θ = opposite side/hypotenuse
Cos 12 = y/8000
x = 8000Cos12 = 8000 × 0.9781
x = 7824.8 feet
In Professor Krugman's economics course, the correlation between the students' total scores prior to the final examination and their final-examination scores is r = 0.5. The pre-exam totals for all students in the course have mean 280 and standard deviation 40. The final-exam scores have mean 75 and standard deviation 8. Professor Krugman has lost Julie's final exam but knows that her total before the exam was 300. He decides to predict her final-exam score from her pre-exam total.
Which of the choices is the least-squares regression line of final-exam scores on pre-exam total scores in this course?
(A) y =-0.1-47x
(B) y=47-0.1x
(C) y=-47 + 0.1x
(D) y= 47 + 0.1x
Answer:
D
Step-by-step explanation:
The regression equation is
y=a+bx
Now, we have to estimate a and b.
We know that slope can be determine as
b=byx=r*(sy/sx)
=0.5*(8/40)
b=0.1
So, we have the slope of regression equation is 0.1.
Now the intercept "a" can be estimated as
y=a+bx
a=ybar-bxbar
a=75-0.1*280
a=47
So, we have the intercept of regression equation is 47.
y=a+bx
y=47+0.1x
Final answer:
The least-squares regression line of final-exam scores on pre-exam total scores, given the statistics provided, is y = 47 + 0.1x, which corresponds to option (D).
Explanation:
The question involves identifying the least-squares regression line of final-exam scores on pre-exam total scores in an economics course. Given the correlation coefficient (r = 0.5), the mean and standard deviation of both the pre-exam totals and the final-exam scores, we can use the formula for the regression line, which is:
y = b0 + b1*x
where y is the predicted final exam score, x is the pre-exam total score, b0 is the y-intercept, and b1 is the slope of the line. The slope (b1) is calculated by:
b1 = r * (standard deviation of y / standard deviation of x) = 0.5 * (8 / 40) = 0.1
And the y-intercept (b0) is found using the means of x and y:
b0 = mean of y - b1 * mean of x = 75 - 0.1 * 280 = 47
So, the regression line is y = 47 + 0.1x, which corresponds to option (D).