Answer:
B. Yes, it is equivalent
Step-by-step explanation:
A = (-3, 2, 3)
B = (-3, 5, 2)
/AB/ = (-3-(-3), 2-5, 3-2)
= (0, -3, 1)
P = (2, -3, 2)
Q = (2, 0, 1)
/PQ/ = (2-2, -3-0, 2-1)
= (0, -3, 1)
So, /AB/ is equivalent to /PQ/
On any given day, the probability that a large local river is polluted by carbon tetrachloride is 0.12. Each day, a test is conducted to determine whether the river is polluted by carbon tetrachloride. This test has proved correct 84 percent of the time.1. What proportion of days does the test indicate carbon tetrachloride pollution? 2. Suppose that on a particular day the test indicated carbon tetrachloride pollution What is the probability that such pollution actually exists? 3. What percentage of days where the test is negative will the river actually be polluted?
Answer:
a) 0.2416
b) 0.4172
c) 0.0253
Step-by-step explanation:
Since the result of the test should be independent of the time , then the that the test number of times that test proves correct is independent of the days the river is correct .
denoting event a A=the test proves correct and B=the river is polluted
a) the test indicates pollution when
- the river is polluted and the test is correct
- the river is not polluted and the test fails
then
P(test indicates pollution)= P(A)*P(B)+ (1-P(A))*(1-P(B)) = 0.12*0.84+0.88*0.16 = 0.2416
b) according to Bayes
P(A∩B)= P(A/B)*P(B) → P(A/B)=P(A∩B)/P(B)
then
P(pollution exists/test indicates pollution)=P(A∩B)/P(B) = 0.84*0.12 / 0.2416 = 0.4172
c) since
P(test indicates no pollution)= P(A)*(1-P(B))+ (1-P(A))*P(B) = 0.84*0.88+ 0.16*0.12 = 0.7584
the rate of false positives is
P(river is polluted/test indicates no pollution) = 0.12*0.16 / 0.7584 = 0.0253
Cody and Chelsea are 270 feet apart when they start walking toward one another. They are walking at the same speed, so whenever Cody travels some number of feet, Chelsea travels the same number of feet. Let x represent the number of feet Cody has traveled since he started walking toward Chelsea.
Write an expression in terms of xx that represents the number of feet Cody has walked toward Chelsea since they started walking.
The expression in terms of [tex]x[/tex] that represents the number of feet Cody has walked toward Chelsea since they started walking is [tex]x_{Co} = x[/tex].
Procedure - Uniform motionIn this question we assume that both Cody and Chelsea have a uniform motion, in opposite sides and to their encounter. If both travel the same number of feet, the kinematic formulas for each walker are described below:
Cody[tex]x_{Co} = x[/tex] (1)
Chelsea[tex]x_{Ch} = 270 - x[/tex] (2)
Encounter[tex]x_{Co} = x_{Ch}[/tex] (3)
Where:
[tex]x[/tex] - Number of feet walked by each walker, in feet.[tex]x_{Co}[/tex], [tex]x_{Ch}[/tex] - Position of each walker, in feet.By (3), (2) and (1) we have the following equation:
[tex]x = 270 - x[/tex]
[tex]2\cdot x = 270[/tex]
[tex]x = 135\,ft[/tex]
The expression in terms of [tex]x[/tex] that represents the number of feet Cody has walked toward Chelsea since they started walking is [tex]x_{Co} = x[/tex]. [tex]\blacksquare[/tex]
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The expression to represent the number of feet Cody has walked toward Chelsea since they started walking is '270 - 2x'. Where 'x' represents the distance Cody has traveled and '2x' reflecting that Chelsea has traveled the same distance. The total is subtracted from 270, the initial distance, to find the remaining distance.
Explanation:Given that Cody and Chelsea start walking towards each other from a distance of 270 feet and that they are walking at the same speed. If x represents the number of feet Cody has traveled since he started walking, given that Chelsea travels the same distance, the total distance they have both traveled together is represented by 2x. Since they started 270 feet apart, to find out how much more they still have to walk until they meet, you would subtract the total feet they've walked from the original distance. So this can be represented by the expression: 270 - 2x. This expression will give us the remaining distance after Cody has walked 'x' feet towards Chelsea.
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Assume the radius of an atom, which can be represented as a hard sphere, is r 1.95 Å.The atom is placed in a ( a ) simple cubic, ( b ) fcc, ( c ) bcc, and ( d ) diamond lattice. As-suming that nearest atoms are touching each other, what is the lattice constant of eachlattice
Answer: Simple cubic=0.39nm
Face centred cubic
=0.55nm
Body centered cubic
=0.45nm
Diamond lattice= 0.9nm
Step-by-step explanation: The lattice constant (a)
for SC=2*r
Fcc=4*r/√2
Bcc= 4*r/√3
Diamond lattice=8*r/√3
Here,
r is the atomic radius measured in nm
r = 1.95Å * 1nm/10Å
=0.195nm
Now let's calculate (a)
SC = 2*r = 2*0.195 nm=0.39nm
Fcc = 4*r/√2 =4*0.195nm/√2
= 0.55nm
Bcc = 4*r/√3 =4*0.195nm/√3
= 0.45nm
Diamond lattice = 8*r/√3
=8*0.195nm/√3
= 0.9nm
What size sample should we draw from a population with standard deviation of 15 to have a sampling distribution standard error of 1.875?1564288
Answer: n = 64
Step-by-step explanation:
Standard error (SE) of a statistical data is the standard deviation of its sampling distribution.
S.E = r/√n. .....1
Where,
SE = standard error
r = standard deviation
n = number of samples
Given:
r = 15
SE = 1.875
From equation 1, making n the subject of formula:
n = (r/SE)^2
n = (15/1.875)
n = 64
In a recent poll of 1100 randomly selected home delivery truck drivers, 26% said they had encountered an aggressive dog on the job at least once. What is the standard error for the estimate of the proportion of all home delivery truck drivers who have encountered an aggressive dog on the job at least once? Round to the nearest ten-thousandth a) 0.0141 b) 0.0002 c) 0.1322 d) 0.0132
Answer:
d) 0.0132
Step-by-step explanation:
In a sample of size n, with proportion p, the standard error of the proportion is:
[tex]SE_{p} = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem, we have that:
[tex]p = 0.26, n = 1100[/tex]
So
[tex]SE_{p} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.26*0.74}{1100}} = 0.0132[/tex]
So the correct answer is:
d) 0.0132
Final answer:
Explaining the calculation of the standard error for a proportion estimate based on a poll of truck drivers encountering aggressive dogs.
Explanation:
Question: In a recent poll of 1100 randomly selected home delivery truck drivers, 26% said they had encountered an aggressive dog on the job at least once. What is the standard error for the estimate of the proportion of all home delivery truck drivers who have encountered an aggressive dog on the job at least once?
Answer:
Calculate the standard error using the formula: SE = sqrt((p*(1-p))/n)
Substitute the values: p = 0.26 (proportion), n = 1100 (sample size)
SE = sqrt((0.26*(1-0.26))/1100) = sqrt((0.26*0.74)/1100) = sqrt(0.1924/1100) = sqrt(0.000175) ≈ 0.0132 (option d)
Given the following data: X 1 4 6 7 Y 9 7 8 1 a) Find the coefficient of correlation. b) Find the equation of the regression line. c) What is the predicted value for Y, if X = 3?
a. [tex]\[ r \approx -0.736 \][/tex]
b. The equation of the regression line is [tex]\(Y = 11.075 - 1.127X\)[/tex].
c. The predicted value for [tex]\(Y\)[/tex] when [tex]\(X = 3\)[/tex] is approximately 7.794.
To find the coefficient of correlation [tex](\(r\))[/tex] and the equation of the regression line, we can follow these steps:
a) Find the Coefficient of Correlation [tex](\(r\))[/tex]:
The formula for the coefficient of correlation [tex](\(r\))[/tex] is given by:
[tex]\[ r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \][/tex]
where \(n\) is the number of data points, [tex]\(\sum xy\)[/tex] is the sum of the product of corresponding values of [tex]\(X\)[/tex] and [tex]\(Y\), \(\sum x\)[/tex] is the sum of [tex]\(X\)[/tex], and [tex]\(\sum y\)[/tex] is the sum of [tex]\(Y\)[/tex].
[tex]\[ n = 4 \][/tex]
[tex]\[ \sum x = 1 + 4 + 6 + 7 = 18 \][/tex]
[tex]\[ \sum y = 9 + 7 + 8 + 1 = 25 \][/tex]
[tex]\[ \sum xy = (1 \times 9) + (4 \times 7) + (6 \times 8) + (7 \times 1) = 9 + 28 + 48 + 7 = 92 \][/tex]
[tex]\[ \sum x^2 = 1^2 + 4^2 + 6^2 + 7^2 = 1 + 16 + 36 + 49 = 102 \][/tex]
[tex]\[ \sum y^2 = 9^2 + 7^2 + 8^2 + 1^2 = 81 + 49 + 64 + 1 = 195 \][/tex]
Now, substitute these values into the formula:
[tex]\[ r = \frac{4 \times 92 - 18 \times 25}{\sqrt{[4 \times 102 - (18)^2][4 \times 195 - (25)^2]}} \][/tex]
[tex]\[ r = \frac{368 - 450}{\sqrt{[408 - 324][780 - 625]}} \][/tex]
[tex]\[ r = \frac{-82}{\sqrt{84 \times 155}} \][/tex]
[tex]\[ r \approx -0.736 \][/tex]
b) Find the Equation of the Regression Line:
The equation of the regression line [tex](\(Y = a + bX\))[/tex] is given by:
[tex]\[ b = r \times \frac{s_y}{s_x} \][/tex]
[tex]\[ a = \bar{y} - b \bar{x} \][/tex]
where [tex]\(s_y\)[/tex] and [tex]\(s_x\)[/tex] are the standard deviations of [tex]\(Y\)[/tex] and [tex]\(X\)[/tex] respectively, and [tex]\(\bar{y}\)[/tex] and [tex]\(\bar{x}\)[/tex] are the means of [tex]\(Y\)[/tex] and [tex]\(X\)[/tex] respectively.
[tex]\[ s_x = \sqrt{\frac{\sum (x - \bar{x})^2}{n-1}} \][/tex]
[tex]\[ s_y = \sqrt{\frac{\sum (y - \bar{y})^2}{n-1}} \][/tex]
[tex]\[ \bar{x} = \frac{\sum x}{n} \][/tex]
[tex]\[ \bar{y} = \frac{\sum y}{n} \][/tex]
[tex]\[ s_x \approx 2.160 \][/tex]
[tex]\[ s_y \approx 3.299 \][/tex]
Now, calculate [tex]\(b\)[/tex] and [tex]\(a\)[/tex]:
[tex]\[ b = -0.736 \times \frac{3.299}{2.160} \][/tex]
[tex]\[ b \approx -1.127 \][/tex]
[tex]\[ \bar{x} = \frac{18}{4} = 4.5 \][/tex]
[tex]\[ \bar{y} = \frac{25}{4} = 6.25 \][/tex]
[tex]\[ a = 6.25 - (-1.127 \times 4.5) \][/tex]
[tex]\[ a \approx 11.075 \][/tex]
So, the equation of the regression line is [tex]\(Y = 11.075 - 1.127X\)[/tex].
c) Predicted Value for Y when X = 3:
[tex]\[ Y = 11.075 - 1.127 \times 3 \][/tex]
[tex]\[ Y \approx 7.794 \][/tex]
So, the predicted value for [tex]\(Y\)[/tex] when [tex]\(X = 3\)[/tex] is approximately 7.794.
Which equation can be used to find the total number of contestants, C that audition in H hours?
Answer:
The correct answer is C. 32h = C
Step-by-step explanation:
For finding out the equation that can be used to find the total number of contestants, C, that audition in h hours. we will use these variables:
Number of contestants = C
Number of hours of auditions = h
Number of contestants that can audition per hour = 32
Now, we can affirm that the equation that can be used to find the total number of contestants, C, that audition in h hours is:
C = 32 * h
The correct answer is C. 32h = C
For what value of x does 3^4 x = 27^x - 3? a.–9 b.–3 c.3 d.9
Answer:
A
Step-by-step explanation:
Which of the following are considered cylinders? Hint : Try to do this as if it were on the exam. Use the traces (or cross-sections) method to draw a picture to help you classify. (Click all that apply to get credit.)
A. 5y^{2}+2z^{2}=2
B. y=-\left(2x^{2}+2\right)
C. 5x^{2}+2y^{2}+4z^{2} = 5
D. 4x^{2}-4y^{2}+4z^{2} = 2
E. -\left(\cos\!\left(-4y\right)\right)=1-z
F. 2x^{2}+4y^{2}-4z = -1
G. None of the above.
Answer:
Step-by-step explanation:
A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given curve in some plane.
a)[tex]5y^{2}+2z^{2} = 2 [/tex]
this is cylinder , this is ellipse in y-z plane ,but along x-axis ,we shall get the same curve in every possible plane parallel to the yz-plane.
b)[tex]y = -2x^{2} - 2 [/tex]
We have to be careful not to draw interpret it as an equation in 2-space even though there is no z - that just means it does not depend upon z. If we sketch this graph in the xy-plane (so z = 0, we obtain the parabola [tex]y = -2x^ 2 - 2[/tex]) . Since there is no z-value in the equation, we shall get the same curve in every possible plane parallel to the x-y-plane. In particular, the graph of this surface will be all vertical lines passing through the curvey = -2x^ 2 - 2 . in the xy-plane. By definition, this makes the graph a cylinder.
C)[tex]5x^{2}+2y^{2}+4z^{2} = 5 [/tex]
this is ellipsoid
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1[/tex]
d)4x^(2)-4y^(2)+4z^(2) = 2
hyperboloid
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1[/tex]
{x,y,z can be exchanged to each other}
e)1-z = - cos (-4y)
this is some curve in y-z plane ,so it is cylinder.
f)
2x^{2}+4y^{2}-4z = -1
elliptical paraboloid
The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ= 3 psi. A random sample of four specimens is tested. The
results are y1=145, y2=153, y3=150 and y4=147.
(a) State the hypotheses that you think should be tested in this experiment
(b) Test these hypotheses using α
= 0.05. What are your conclusions?
(c) Find the
P-value for the test in part (b).
(d) Construct a 95 percent confidence interval on the mean breaking strength.
The 95 percent confidence interval on the mean breaking strength is (143.9764, 153.5236).
We are given that;
Strength for breaking the fiber= 150 psi
σ= 3 psi
Now,
(a) The hypotheses that should be tested in this experiment are:
[tex]$H_0: \mu = 150$[/tex] (The mean breaking strength of the fiber is 150 psi)
[tex]$H_a: \mu < 150$[/tex] (The mean breaking strength of the fiber is less than 150 psi)
This is a one-tailed test, since we are interested in testing whether the mean breaking strength is below a certain value.
(b) To test these hypotheses using [tex]$\alpha = 0.05$[/tex], we need to calculate the test statistic and compare it with the critical value or the p-value. The test statistic is given by:
[tex]$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$[/tex]
where [tex]$\bar{x}$[/tex] is the sample mean, [tex]$\mu_0$[/tex] is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
Plugging in the given values, we get:
[tex]$t = \frac{148.75 - 150}{3/\sqrt{4}} = -1.6667$[/tex]
The critical value for a one-tailed test with[tex]$\alpha = 0.05$[/tex] and n-1 = 3 degrees of freedom is -2.3534, which can be found from a t-table.
Since t > -2.3534 or p > 0.05, we fail to reject the null hypothesis at the 0.05 level of significance. There is not enough evidence to conclude that the mean breaking strength of the fiber is less than 150 psi.
(c) The p-value for the test in part (b) is 0.0834, as mentioned above.
(d) A 95 percent confidence interval on the mean breaking strength is given by:
[tex]$\bar{x} \pm t_{\alpha/2,n-1} \frac{s}{\sqrt{n}}$[/tex]
where [tex]$t_{\alpha/2,n-1}$[/tex] is the critical value for a two-tailed test with [tex]$\alpha = 0.05$[/tex] and n-1 = 3 degrees of freedom, which is 3.1824.
Plugging in the given values, we get:
[tex]$148.75 \pm 3.1824 \frac{3}{\sqrt{4}}$[/tex]
Simplifying, we get:
[tex]$148.75 \pm 4.7736$[/tex]
Therefore, by statistics answer will be (143.9764, 153.5236).
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Let f(x)=x+2 and g(x)=2x+1 .
Graph the functions on the same coordinate plane.
What are the solutions to the equation f(x)=g(x) ?
Enter your answers in the boxes.
Step-by-step explanation:
Given g(x) = x+2 and f(x) = [tex]2^x+1[/tex]
If f(x) = g(x)
⇔[tex]2^x+1[/tex] = x+2
⇔[tex]2^x= x +1[/tex]
Only x= 1 and x=0 will be satisfy the above equation.
If x= 1, it gives and x=0 gives
[tex]2^1= 1+1[/tex] [tex]2^0=0+1[/tex]
⇔2=2 ⇔1=1
Answer:
1 & 0
Step-by-step explanation:
Express the Cartesian coordinates ( - 5,5) in polar coordinates in at least two different ways. Write the point in polar coordinates with an angle in the range 0 < = theta < 2pi (Type an ordered pair. Type an exact answer, using pi as needed.) Write the point in polar coordinates with an angle in the range - 2pi < = theta < 0. [ ] (Type an ordered pair. Type an exact answer. using pi as needed.)
The Cartesian coordinate point (-5,5) can be re-written in polar coordinates in two ways. The polar coordinates equivalent to (-5,5) in the range 0 <= theta < 2pi are (5sqrt(2), 7pi/4) and in the range -2pi <= theta < 0 are (5sqrt(2), -pi/4).
Explanation:The Cartesian coordinate point (-5,5) can be expressed in polar coordinates using the polar coordinates formula: r = sqrt(x^2 + y^2) and theta = arctan(y/x).
For the first part of the question, we are asked to find the polar coordinates where the angle lies in the range 0 <= theta < 2pi. Using the two formulas above, we find r to be sqrt((-5)^2 + 5^2) = sqrt(50) or 5sqrt(2). The angle theta can be calculated using arctan(5/-5) = -pi/4. We add 2pi to this angle because theta should not be negative, and thus obtain a final result of (7pi/4). So, the polar coordinates equivalent to (-5,5) in the range 0 <= theta < 2pi are (5sqrt(2), 7pi/4).
For the second part of the question, we need the polar coordinates where the angle theta lies in the range -2pi <= theta < 0. In this case, the value for r remains the same as before. However, the angle is -pi/4, because it needs to be negative. Hence, the polar coordinates equivalent to (-5, 5) in the range -2pi <= theta < 0 are (5sqrt(2), -pi/4).
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two cards are chosen at random from a deck of 52 playing cards. what is the probability that they are both aces?
Answer:
The probability that they are both aces is 0.00452.
Step-by-step explanation:
Consider the provided information.
Out of 52 playing card we need to select only 2.
Thus, the sample space is: [tex]^{52}C_2=1326[/tex]
Two cards are Ace.
The number of Ace in a pack of playing card are 4 and we need to select two of them.
This can be written as: [tex]^4C_2=6[/tex]
Thus, the probability that they are both aces is:
[tex]P(\text{Both are Ace})=\dfrac{6}{1326} \approx0.00452[/tex]
Hence, the probability that they are both aces is 0.00452.
Harry church borrowed 17,500.00 at 6.5% exact interest. He had to pay back a maturity value of 17,873.97 to pay off the loan. What was the term of the loan
Answer:the term of the loan is approximately 4 months
Step-by-step explanation:
The term of the loan means the period for which the loan was given.
We would apply the formula for simple interest which is expressed as
I = PRT/100
Where
P represents the principal
R represents interest rate
T represents time in years
I = interest after t years
From the information given
P = 17500
R = 6.5%
I = total amount paid - principal
I = 17,873.97 - 17,500.00 = 373.97
Therefore
373.97 = (17500 × 6.5 × T)/100
373.97 = 1137.5T
T = 373.97/1137.5
T = 0.32 years
Converting to months, it becomes
0.32 × 12 = 3.84
Approximately 4 months.
Find an explicit solution of the given initial-value problem.dy/dx = ye^x^2, y(3) = 1.
Answer:
[tex]y=exp(\int\limits^x_4 {e^{-t^{2} } } \, dt)[/tex]
Step-by-step explanation:
This is a separable equation with an initial value i.e. y(3)=1.
Take y from right hand side and divide to left hand side ;Take dx from left hand side and multiply to right hand side:
[tex]\frac{dy}{y} =e^{-x^{2} }dx[/tex]
Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):
[tex]\int\limits^x_3 {\frac{1}{y} } \, \frac{dy}{dt} dt=\int\limits^x_3 {e^{-t^{2} } } dt[/tex]
Integrate on both sides:
[tex]ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^{2} } } \, dt[/tex]
Use initial condition i.e. y(3) = 1:
[tex]ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^{2} } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^{2} } } \, dt\\[/tex]
Taking exponents on both sides to remove "ln":
[tex]y=exp (\int\limits^x_3 {e^{-t^{2} } } \, dt)[/tex]
Consider the following planes. 3x − 2y + z = 1, 2x + y − 3z = 3
Find parametric equations for the line of intersection of the planes.
Solve for any of the variables; for instance, [tex]z[/tex]:
[tex]z=1-3x+2y[/tex]
[tex]z=-\dfrac{3-2x-y}3[/tex]
Then
[tex]1-3x+2y=-\dfrac{3-2x-y}3\implies11x-5y=6\implies\begin{cases}y=\frac{11x-6}5\\z=1-3x+\frac{22x-12}5\end{cases}[/tex]
Let [tex]x=t[/tex]; then the intersection is given by the vector-valued function
[tex]\vec r(t)=\left(t,\dfrac{11t-6}5,1-3t+\dfrac{22t-12}5\right)[/tex]
or
[tex]\vec r(t)=\left(t,\dfrac{11t-6}5,\dfrac{7t-7}5\right)[/tex]
Suppose a manufacturer of light bulbs produces a 75-watt bulb that burns a mean of 7500 hours before it burns out. It has a standard deviation of 220 hours. What percentage of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours?
Answer:
8.69% of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 7500, \sigma = 220[/tex]
What percentage of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours?
This is the probability that X is lower or equal than 7200 hours. So this is the pvalue of Z when X = 7200.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7200 - 7500}{220}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.36[/tex] has a pvalue of 0.0869
So 8.69% of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours.
Final answer:
The manufacturer would need to replace approximately 8.69% of their 75-watt light bulbs under a warranty of 7200 hours, based on the given mean lifetime and standard deviation.
Explanation:
To calculate the percentage of 75-Watt light bulbs that a manufacturer would have to replace under warranty if they offered a warranty of 7200 hours, we need to use the normal distribution. The mean lifetime of the bulbs is 7500 hours with a standard deviation of 220 hours. To find the percentage of bulbs that burn out before 7200 hours, we first calculate the z-score, which is the number of standard deviations an observation is from the mean.
The z-score is calculated as:
Z = (X - μ) / σ
Where μ is the mean, σ is the standard deviation, and X is the value we are interested in (7200 hours in this case).
Plugging in the numbers:
Z = (7200 - 7500) / 220 = -300 / 220 = -1.36
Using a Z-table or a statistical calculator, we find that the probability corresponding to a Z-score of -1.36 is approximately 0.0869 or 8.69%. This means that the manufacturer would have to replace about 8.69% of their bulbs under this warranty.
The mean height of a basketball team is 6 feet with a standard deviation of 0.2 feet. The team's center is 6.7 feet tall. Find the center's z score. Is his score unusual?
Answer:
z-score = 3.5, Yes, z-score is unusual.
Step-by-step explanation:
Given information:
Population mean ; μ = 6 feet
Standard deviation ; σ = 0.2 feet
Sample mean = 6.7 feet
The formula for z-score is
[tex]z=\dfrac{\overline{X}-\mu}{\sigma}[/tex]
where, [tex]\overline{X}[/tex] is sample mean, μ is population mean and σ is standard deviation.
Substitute the given values in the above formula.
[tex]z=\dfrac{6.7-6}{0.2}[/tex]
[tex]z=\dfrac{0.7}{0.2}[/tex]
[tex]z=3.5[/tex]
The z-score is 3.5.
If a z-score is less than -2 or greater than 2, then it is known as unusual score.
3.5 > 2
It means z-score is unusual.
Since the z score is greater than 3, hence his score is unusual.]
Z scoreZ score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma} \\ \\ where\ x=raw\ score,\mu=mean\ and\ \sigma=standard\ deviation[/tex]
Given that mean = 6, standard deviation = 0.2.
For x = 6.7:
[tex]z=\frac{6.7-6}{0.2}=3.5 [/tex]
Since the z score is greater than 3, hence his score is unusual.
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A music school has budgeted to purchase 3 musical instruments. They plan to purchase a piano costing $4,000, a guitar costing $600, and a drum set costing $750. The mean cost for a piano is $4,500 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $850 with a standard deviation of $100. (Enter your answers to two decimal places.)
1.) How many standard deviations above or below the average piano cost is the piano?
2.) How many standard deviations above or below the average guitar cost is the guitar?
3.) How many standard deviations above or below the average drum set cost is the drum set?
4.) Which cost is the lowest, when compared to other instruments of the same type?
5.) Which cost is the highest when compared to other instruments of the same type?
The piano is 0.2 standard deviations below the average piano cost, while the guitar is 0.5 standard deviations above the average guitar cost. The drum set is 1 standard deviation below the average drum set cost. The drums have the lowest cost compared to other instruments of the same type, and the piano has the highest cost.
Explanation:1.) To find how many standard deviations above or below the average piano cost is the piano, we need to calculate the z-score. The formula for calculating the z-score is
z = (x - mean) / standard deviation
Let's plug in the values for the piano:
z = ($4,000 - $4,500) / $2,500
z = -0.2
So, the piano is 0.2 standard deviations below the average piano cost.
2.) Following the same formula, let's calculate the z-score for the guitar:
z = ($600 - $500) / $200
z = 0.5
Thus, the guitar is 0.5 standard deviations above the average guitar cost.
3.) Using the formula for the z-score again, let's calculate the z-score for the drum set:
z = ($750 - $850) / $100
z = -1
The drum set is 1 standard deviation below the average drum set cost.
4.) The cost of the drums is the lowest when compared to other instruments of the same type.
5.) The cost of the piano is the highest when compared to other instruments of the same type.
What is the rate of change of a linear function that includes the points (5, 2) and (9, 4)?
A) -2
B) 2
C) 4
D) 1/2
1
2
Answer:
D) 1/2
Step-by-step explanation:
A linear function has the following format:
[tex]y = ax + b[/tex]
In which a is the slope, that is, the rate of change of the function and b is the y-intercept, that is, the value of y when x = 0.
We have that:
Point (5,2), which means that when x = 5, y = 2. So
[tex]y = ax + b[/tex]
[tex]5a + b = 2[/tex]
Point (9,4), which means that when x = 9, y = 4. So
[tex]9a + b = 4[/tex]
I am going to write b as a function of a in the second equation, and replace on the first. So
[tex]b = 4 - 9a[/tex]
Then
[tex]5a + b = 2[/tex]
[tex]5a + 4 - 9a = 2[/tex]
[tex]-4a = -2[/tex]
[tex]4a = 2[/tex]
[tex]a = 1/2[/tex]
This means that the slope, which is the rate of change of the function, is 1/2.
So the correct answer is:
D) 1/2
In late June 2012, Survey USA published results of a survey stating that 56% of the 600 randomly sampled Kansas residents planned to set off fireworks on July 4th. Determine the margin of error for the 56% point estimate using a 95% confidence level.
Answer:
The margin of error is of 3.97 percentage points.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The confidence interval has the following margin of error
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
We also have that
[tex]n = 600, \pi = 0.56[/tex]
So, the margin of error is:
[tex]M = 1.96*\sqrt{\frac{0.56*0.44}{600}} = 0.0397[/tex]
So the margin of error is of 3.97 percentage points.
The margin of error for the 56% point estimate using a 95% confidence level is; 3.97%
Formula for margin of error here is;
M = z√(p(1 - p)/n)
Where;
z is critical value at given confidence level
p is sample proportion
n is sample size
We are given;
p = 56% = 0.56
n = 600
Confidence level = 95%
Now,from tables, the critical value at a confidence level of 95% is; z = 1.96
Thus;
M = 1.96√(0.56(1 - 0.56)/600)
M = 0.0397 or 3.97%
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Express the confidence interval (0.403 comma 0.487 )(0.403,0.487) in the form of ModifyingAbove p with caretpplus or minus±E. ModifyingAbove p with caretpplus or minus±Eequals=nothingplus or minus±nothing (Type an integer or a decimal.)
L = 0.403 is the lower end of the confidence interval
U = 0.487 is the upper end of the confidence interval
The margin of error E is
E = (U-L)/2
E = (0.487-0.403)/2
E = 0.042
Which is half the distance from the lower to upper end
---------------------------
The midpoint of the confidence interval is the value of [tex]\hat{p}[/tex] (read out as "p hat")
phat = (U+L)/2
phat = (0.487+0.403)/2
phat = 0.445
---------------------------
Answer: The confidence interval (0.403, 0.487) can be rewritten into the form [tex]0.445 \pm 0.042[/tex]
The correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
What fraction of the variability in fuel economy is accounted for by the engine size?
A. 27.5%
B. 63.8%
C. 71.8%
D. 84.1%
Answer: 59.91%.
Step-by-step explanation:
We know that the fraction of the variability in data values accounted by a model is given by [tex]r^2[/tex] , where r is the coefficient of correlation.We are given , that the correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
Then, the fraction of the variability in fuel economy is accounted for by the engine size would be [tex]r^2=( - 0.774)^2=0.599076\approx59.91\%[/tex]
[Multiply 100 to convert a decimal into percent]
Hence, the fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
None of the options are correct.
To solve such problems we must know about the fraction of the variability in data values or R-squared.
R- SquaredThe fraction by which the variance of the dependent variable is greater than the variance of the errors is known as R-squared. It is called so because it is the square of the correlation between the dependent and independent variables, which is commonly denoted by “r” in a simple regression model.fraction of the variability in data values = (coefficient of correlation)²= r²
The fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
Given to usThe correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.SolutionVariability in fuel economyAs it is given that the correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
the variability in fuel economy = r²
= (-0.774)²
= 0.599076
= 59.91%
Hence, the fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
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a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.79 million dollars. Assuming a population standard deviation gross earnings of 0.47 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: ( __________________ , __________________ )
Answer:
The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.5690, 3.0110).
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{0.47}{\sqrt{30}} = 0.2210[/tex]
The lower end of the interval is the mean subtracted by M. So it is 2.79 - 0.2210 = 2.5690 million dollars.
The upper end of the interval is the mean added to M. So it is So it is 2.79 + 0.2210 = 3.0110 million dollars.
The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.5690, 3.0110).
The average age of students in a statistics class is 28 years. The 28 years would be considered an example of:
Based on the information given, it should be noted that the 28 years would be considered an example of descriptive statistics.
It should be noted that descriptive statistics gives a summary description of a particular data quantitatively.It is given to an analysis of data which is vital in describing or showing the summary of data.Therefore, 28 years would be considered an example of descriptive statistics.
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The average age of 28 years in the given class is an example of a statistic, which is a numerical value calculated from a sample of a larger population.
In the context of this question from the subject of mathematics, specifically statistics, the average age of 28 years for students in a class is considered an example of a statistic. A statistic is a value calculated from a sample drawn from a larger population. In this case, the sample consists of students in this particular class, and the population could be all students taking a statistics class in the country or the world. The average (or mean) is a basic type of statistic used to summarise a set of values. It's important to note that the average is just one type of statistic - there are many others, such as median, mode, range, and standard deviation that provide different ways to summarise and represent data.
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Find the area of the region bounded by the hyperbola 25x2 − 4y2 = 100 and the line x = 3. (Using trigonometric substitution)
The question involves calculating the area between a hyperbola and a line using trigonometric substitution, yet it's more directly related to understanding hyperbola properties and standard integration techniques of hyperbolic functions rather than a straightforward trigonometric substitution problem.
Explanation:The problem involves finding the area bounded by the hyperbola 25x² - 4y² = 100 and the line x = 3 using trigonometric substitution. This specific problem requires a solid understanding of calculus, particularly integration, and knowledge of hyperbolas and trigonometric identities. To solve this, we first rewrite the hyperbola in its standard form, which is √(25x²/100 - 4y²/100) = 1, simplifying to x²/4 - y²/25 = 1, indicating that a = 2 and b = 5. Using trigonometric substitution, we set x = a sec(θ) and y = b tan(θ) where a and b are the semi-major and semi-minor axes of the hyperbola, respectively. However, the given question simplifies to the computation of an area for a given range of x, particularly up to x = 3, which doesn't directly imply a need for integral computation but rather an understanding of the geometry of hyperbolas.
The straight line x = 3 intersects the hyperbola at specific points, and the area in question could be typically found using trigonometric integrals if it required finding an area under a curve or between two curves. In this context, without further details on the exact method of trigonometric substitution to be used for computing the area directly, we often rely on the integrals involving the hyperbolic trigonometric functions, using limits set by the intersections of the curve and the line. Since a direct solution involving trigonometric substitution for area calculation under these conditions is complex and not standard, it's crucial to review the application within this specific problem's context further.
Laura and Phillip each fire one shot at a target, Laura has aprobability 0.5 of hitting the target, and Phillip has aprobability 0.3. The shots are independent.
a. Find the probability that the target is hit.
b.Find the probability that the target is hit by exactly oneshot.
c.Given that the target was hit by exactly one shot, find theprobabilty that Laura hit the target.
a. The probability that the target is hit is 0.65.
b. The probability that the target is hit by exactly one shot is 0.5.
c. Given exactly one shot hits, the probability Laura hit is 0.7.
a. To find the probability that the target is hit, we can calculate the probability that at least one shot hits the target. Since the shots are independent, we can use the complement rule:
[tex]\[ P(\text{Target hit}) = 1 - P(\text{Both miss}) \][/tex]
Laura misses with probability 0.5 and Phillip misses with probability 0.7 (since the probability of hitting the target is 0.3). Therefore:
[tex]\[ P(\text{Both miss}) = 0.5 \times 0.7 = 0.35 \][/tex]
[tex]\[ P(\text{Target hit}) = 1 - 0.35 = 0.65 \][/tex]
So, the probability that the target is hit is 0.65.
b. To find the probability that the target is hit by exactly one shot, we need to consider the cases where Laura hits and Phillip misses, and where Laura misses and Phillip hits. These events are mutually exclusive, so we can add their probabilities:
[tex]\[ P(\text{Exactly one shot hits}) = P(\text{Laura hits, Phillip misses}) + P(\text{Laura misses, Phillip hits}) \][/tex]
[tex]\[ = (0.5 \times 0.7) + (0.5 \times 0.3) \][/tex]
= 0.35 + 0.15
= 0.5
So, the probability that the target is hit by exactly one shot is 0.5.
c. Given that the target was hit by exactly one shot, we need to find the probability that Laura hit the target. This is the conditional probability:
[tex]\[ P(\text{Laura hit} | \text{Exactly one shot hit}) = \frac{P(\text{Laura hit and exactly one shot hit})}{P(\text{Exactly one shot hit})} \][/tex]
From part b, we found [tex]\( P(\text{Exactly one shot hit}) = 0.5 \)[/tex]. And from part b, we found [tex]\( P(\text{Laura hit and exactly one shot hit}) = 0.35 \)[/tex].
[tex]\[ P(\text{Laura hit} | \text{Exactly one shot hit}) = \frac{0.35}{0.5} = 0.7 \][/tex]
So, the probability that Laura hit the target, given that the target was hit by exactly one shot, is 0.7.
there is a herd of cattle out on the range. fifty-two percent of the cattle are male. if the herd consists of 1175 animals, how many are female
Answer:
564 of the cattle are female
Step-by-step explanation:
if 52% of them are male, then 48% are female.
.48*1175=564
Answer:
1227
Step-by-step explanation:
You research commute times to work and find that the population standard deviation is 9.3 minutes. Repeat Exercise, using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. In a random sample of eight people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.2 minutes.
Answer:
Case [tex] s =7.2[/tex]
[tex]35.5-2.36\frac{7.2}{\sqrt{8}}=29.49[/tex]
[tex]35.5+2.36\frac{7.2}{\sqrt{8}}=41.51[/tex]
So on this case the 95% confidence interval would be given by (29.49;41.51)
Case [tex] \sigma =9.3[/tex]
[tex]35.5-1.96\frac{9.3}{\sqrt{8}}=29.06[/tex]
[tex]35.5+1.96\frac{9.3}{\sqrt{8}}=41.94[/tex]
So on this case the 95% confidence interval would be given by (29.06;41.94)
And we conclude that the intervals are very similar.
Step-by-step explanation:
If we assume that for this question we need to find a confidence interval for the population mean. We have the following procedure:
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= 35.5[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=7.2 represent the sample standard deviation
n=8 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=8-1=7[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that [tex]t_{\alpha/2}=2.36[/tex]
Now we have everything in order to replace into formula (1):
[tex]35.5-2.36\frac{7.2}{\sqrt{8}}=29.49[/tex]
[tex]35.5+2.36\frac{7.2}{\sqrt{8}}=41.51[/tex]
So on this case the 95% confidence interval would be given by (29.49;41.51)
If we assume that the real population standard deviation is [tex] \sigma =9.3[/tex] the confidence interval is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
[tex]35.5-1.96\frac{9.3}{\sqrt{8}}=29.06[/tex]
[tex]35.5+1.96\frac{9.3}{\sqrt{8}}=41.94[/tex]
So on this case the 95% confidence interval would be given by (29.06;41.94)
And we conclude that the intervals are very similar.
Sketch the region of integration for the following integral. ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ
Answer:
The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
Step-by-step explanation:
We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
The region of integration is a rectangular region in the polar coordinate system, bounded by the angles 0 and π/40, and the radii 0 and 6/cos(θ).
Explanation:The region of integration for the given integral is a rectangular region in the polar coordinate system, bounded by the angles 0 and π/40, and the radii 0 and 6/cos(θ).
To sketch this region, draw a rectangle in the polar coordinate plane, where the horizontal sides represent the radii and the vertical sides represent the angles. The bottom of the rectangle is at radius 0 and the top is at radius 6/cos(θ). The left side of the rectangle is at angle 0 and the right side is at angle π/40.
Therefore, the region of integration can be represented by the rectangle with sides defined by the radii 0 and 6/cos(θ), and the angles 0 and π/40.