Answer:
option C. (x + 4) and (x − 8)
Step-by-step explanation:
A factor is one of the linear expressions of a single-variable of the polynomial.
Given: y = (1/4)(x + 4)(x - 8)
When y = 0
∴ (1/4)(x + 4)(x - 8) = 0 ⇒ multiply both sides by 4
∴ (x + 4)(x - 8) = 0
So, the factors of the function are (x+4) and (x-8)
The answer is option C. (x + 4) and (x − 8)
Which of the following points represents -2 1/4?
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
Find the third Taylor polynomial P3(x) for the function f(x) = (x-1)ln(x) about x0 = 1.
(a) Use P3(0.5) to approximate f(0.5). Find an upper bound for error |f(0.5) - P3(0.5)| using the error formula, and compare it to the actual error.
(b) Find a bound for the error |f(x) - P3(x)| in using P3(x) to approximate f(x) on the interval [0.5,1.5].
(c) Approximate \int_{0.5}^{1.5}f(x)dx using \int_{0.5}^{1.5}P_{3}(x)dx .
(d) Find an upper bound for the error in (c) using \int_{0.5}^{1.5}|R_{3}(x)dx| , and compare the bound to the actual error.
Answer:
A) The actual error |f(0.5) - P3(0.5)| is 0.0340735. The error using Taylor's error formula R₃(0.5) is 0.2917.
B) Te error in the interval [0.5,1.5] is 0.0340735.
C) The approximate integral using P₃(x) is 1/12
D) The real error between the two integrals is 4.687·10^(-3) while the error using \int_{0.5}^{1.5}|R_{3}(x)dx| is 0.018595
Step-by-step explanation:
A) To determine the error we first have to write the third grade Taylor polynomial P3(x):
[tex]P_n(x)=\displaystyle\sum_{k=0}^N \frac{f^{k}(a)}{k!}(x-a)[/tex] with N=3
[tex]P_3(x)=\displaystyle\sum_{k=0}^3 \frac{f^{k}(1)}{k!}(x-1)=(x-1)^2-\frac{1}{2} (x-1)^3[/tex]
The error for the third grade Taylor polynomial R₃(x) is represented as the next term non-written in the polynomial expression near to the point x:
[tex]R_3(x)=\displaystyle \frac{f^{4}(x)}{4!}(x-a)=\frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4[/tex]
Therefore the real error is:
[tex]Err(0.5)=|f(0.5)-P_3(0.5)|=0.0340735[/tex]
The error of the Taylor polynomial R₃(0.5) is:
[tex]R_3(0.5)=\displaystyle \frac{1}{24} (\frac{2}{0.5^3}+\frac{6}{0.5^4})(0.5-1)^4=0.2917[/tex]
B) The error in the interval [0.5,1.5] is the maximum error in that interval.
This is found in the extremes of the intervals. We analyze what happens in X=1.5:
[tex]Err(1.5)=|f(1.5)-P_3(1.5)|=0.01523[/tex]
[tex]R_3(1.5)=\displaystyle \frac{1}{24} (\frac{2}{1.5^3}+\frac{6}{1.5^4})(1.5-1)^4=\frac{1}{216}=4.63\cdot 10^{-3}[/tex]
Both of these errors are smaller than Err(0.5) and R₃(0.5). Therefore the error in this interval is Err[0.5,1.5] is Err(0.5).
C) The approximation of the integral and the real integral is:
[tex]I_f=\displaystyle\int_{0.5}^{1.5} f(x)\, dx=\int_{0.5}^{1.5} (x-1)ln(x)\, dx=0.08802039[/tex]
[tex]I_{p3}=\displaystyle\int_{0.5}^{1.5} P_3(x)\, dx=\int_{0.5}^{1.5} (x-1)^2-0.5(x-1)^3\, dx=1/12=0.0833333[/tex]
D) The error in the integrals is:
[tex]Errint=|I_f-I_{p3}|=4.687\cdot10^{-3}[/tex]
[tex]Errint(R_3)=\displaystyle\int_{0.5}^{1.5} R_3(x)\, dx=\int_{0.5}^{1.5} \frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4\, dx=0.018595[/tex]
A mathematician tells her son, "I will give you $100 only if you pass the class." Being a mathematician, she means exactly what she said. Check all true statements about this situation. a. If she ended up paying him $100, he must have passed the class. b. Her statement defined passing the class as a necessary condition for receiving the $100. c. if he passes the class, she will pay him $100. d. Her statement defined passing the class as a sufficient condition for receiving the 100 e. He will not get the s 100 unless he passes the class. f. If she didn't pay him $100, that means he didnt pass the class. . g. If he doesn't pass the class, she will definitely not pay him $100. h. If he passes the class, she may pay him $100.
Answer: C. If he passes the class, she will pay $100.
A. If she ended up paying $100, he must have passed the class.
Step-by-step explanation:
For a statement to be true, one assumption might imply the other assumption. In this case, A and C fulfills it.
Final answer:
The mathematician's statement makes passing the class a necessary condition for her son to receive $100. If he received the money, he must have passed the class, but not receiving the money does not conclusively prove he did not pass. The statement does not make passing the class a sufficient condition, as there could be other requirements.
Explanation:
The statement made by the mathematician to her son is a classic example of a conditional statement in logic, particularly looking at necessary and sufficient conditions. To analyze the truth of the given statements, we must understand the relationship between passing the class and receiving the $100.
a. If she ended up paying him $100, he must have passed the class. This is true. The payment is made on the condition of passing the class, so if he got paid, it means the condition was met.b. Her statement defined passing the class as a necessary condition for receiving the $100. This is true. 'Necessary condition' means without passing the class, he cannot get the $100.c. If he passes the class, she will pay him $100. This seems true based on the statement; however, we do not know if there are other conditions or possibilities so this could be false.d. Her statement defined passing the class as a sufficient condition for receiving the $100 False. Her statement does not guarantee that passing the class is all that is required for him to receive the $100; there might be other conditions.e. He will not get the $100 unless he passes the class. True. This is another way of saying that passing the class is a necessary condition.f. If she didn't pay him $100, that means he didn't pass the class. False. Not receiving the $100 could also occur due to other conditions not being met.g. If he doesn't pass the class, she will definitely not pay him $100. True. This is the definition of a necessary condition for the payment being made.h. If he passes the class, she may pay him $100. True. This acknowledges the possibility of passing the class leading to payment but does not imply it is a guarantee.help pls asap in Ixl for geometry
Answer:
Step-by-step explanation:
Triangle QRS is a right angle triangle.
From the given right angle triangle
RS represents the hypotenuse of the right angle triangle.
With 30 degrees as the reference angle,
QR represents the adjacent side of the right angle triangle.
QS represents the opposite side of the right angle triangle.
To determine QR, we would apply trigonometric ratio
Cos θ = adjacent side/hypotenuse side. Therefore,
Cos 30 = QR/14
√3/2 = QR/14
QR = 14 × √3/2
QR = 14√3/2 = 7√3/2
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).
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
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
Lisa put tulips and roses into vases. Each vase has 12 flowers. The red vase has 7 tulips. the blue vase has twice as mnay roses as the red vase. How many roses are in the blue vase?
Answer:
10 roses
Step-by-step explanation:
To find the number of roses in the blue vase, we double the number of roses in the red vase. Since the red vase has 5 roses, the blue vase contains 10 roses.
Explanation:The student's question is about determining the number of roses in the blue vase given certain conditions about the number of flowers in vases. We know that each vase has 12 flowers, the red vase has 7 tulips, and thus 5 roses because the total is 12. The blue vase has twice as many roses as the red vase, so we need to double the number of roses in the red vase to find that number.
Since the red vase has 5 roses, the blue vase will have twice as many, which is 10 roses.
Here's the calculation step by step:
Subtract the number of tulips from the total flowers in the red vase to find the number of roses: 12 flowers - 7 tulips = 5 roses.Multiply the number of roses in the red vase by 2 to find the number in the blue vase: 5 roses x 2 = 10 roses.A population model for two species can describe either competition for resources (an increase in one species decreases the growth rate in the other) or cooperation for resources (an increase in one species increases the growth rate in the other) For the systems below identify the parameters as growth rates, carrying capacities, and measure of interactions between species. Determine if the species compete of cooperate r, = 31-334-5ry 50 y (a) Growth Rate Constants (b) Carrying Capacity Constants: (c) Interaction Constants: Note: Your answers must be numbers, (which may include 0). If there is more than one answer, separate your answers with commas (d) Choose One Note: You can eam partial credit on this probiem.
(a) Growth Rate Constants: 6
(b) Carrying Capacity Constants: 0
Interaction Constants: 3
The species compete since the interaction constant is positive, indicating competition for resources between the two species.
For the given system:
[tex]\[ x' = 6x \][/tex]
[tex]\[ S_{xy} = 32y^2 - 8y - 8 + 3xy - 31 \][/tex]
(a) Growth Rate Constants: The growth rate constant for species [tex]\(x\)[/tex] is [tex]\(6\)[/tex].
(b) Carrying Capacity Constants: There is no explicit term indicating a carrying capacity for either species in the given equations, so the carrying capacity constants are both [tex]\(0\)[/tex].
Interaction Constants: The interaction constant between the two species can be identified from the term [tex]\(S_{xy}\)[/tex], which is [tex]\(3\)[/tex] in this case.
So, the answers are:
(a) Growth Rate Constants: [tex]\(6\)[/tex]
(b) Carrying Capacity Constants: [tex]\(0\)[/tex]
Interaction Constants: [tex]\(3\)[/tex]
The species compete since the interaction constant is positive.
Complete Question :
Population models for two species can report either competition for resources (an rise in one species recede the growth rate in the other) or cooperation for asset (an rise in one species rise the growth rate in the other). For the systems below, identify the boundary as growth rates, carrying capacities, and measures of interconnection between species. Determine if the species compete or cooperate. x' = 6x Sxy = 32y^2 - 8y - 8 + 3xy - 31 (a) Growth Rate Constants: (b) Carrying Capacity Constants: interconnection Constants: Note: Your answers must be numbers (which may include 0). If there is more than one answer, different your answers with commas. Choose One: - The Species Do Not Interact - One species take part, the other Cooperates - Help Both Species Cooperate - Both Species Compete.
Listed below are the annual tuition amounts of the 10 most expensive colleges in a country for a recent year.
What does this "Top 10" list tell about the population of all that country's college tuitions?
Find the mean, midrange, median, and mode of the data set. Round to two decimal places as needed.
$52,384 $54,079 $52,762 $54,364 $52,584
$53,987 $53,108 $51,588 $50,554 $53,987
Answer:
[tex]\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7[/tex]
[tex] Mode= 53987[/tex]
[tex] Median = \frac{52762+53108}{2}=52935[/tex]
[tex] Midrange=\frac{50554+54364}{2}=52459[/tex]
Step-by-step explanation:
We can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And if we replace the values given we got:
[tex]\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7[/tex]
The mode is the most repeated value and for this case with a frequency of 2 the mode is:
[tex] Mode= 53987[/tex]
In order to find the median we need to order the dataset on increasing way like this:
50554, 51588, 52384 ,52584, 52762
53108, 53987 , 53987 , 54079, 54364
Since we have 10 values an even number the median is calculated from the average between positions 5 and 6 on the data ordered, and we got:
[tex] Median = \frac{52762+53108}{2}=52935[/tex]
The mid range is defined like this:
[tex] Midrange = \frac{Max +Min}{2}[/tex]
And if we replace we got:
[tex] Midrange=\frac{50554+54364}{2}=52459[/tex]
The 'Top 10' list gives insights into the population of college tuitions in a country. The mean, midrange, median, and mode of the data set are $53,138.50, $52,459, $52,584., and $53,987 respectively.
The "Top 10" list of annual tuition amounts for the most expensive colleges in a country provides insights into the range and distribution of college tuitions within the country's higher education system.
Mean: The mean (average) tuition is calculated by summing up all the tuition amounts and dividing by the total number of colleges (in this case, 10).
Mean = ($52,384 + $54,079 + $52,762 + $54,364 + $52,584 + $53,987 + $53,108 + $51,588 + $50,554 + $53,987) / 10 = $53,138.50 (rounded to two decimal places).
Midrange: The midrange is the average of the minimum and maximum values in the dataset.
Midrange = ($50,554 + $54,364) / 2 = $52,459 (rounded to two decimal places).
Median: The median is the middle value when the data is arranged in ascending or descending order. In this case, the data is already provided in ascending order, so the median is the fifth value.
Median = $52,584.
Mode: The mode is the value that appears most frequently in the dataset.
Mode = $53,987.
In summary, the "Top 10" list of college tuitions in the country indicates that there is a range of tuition amounts, with the mean being around $53,138.50. The midrange falls around $52,459, the median is $52,584, and the mode is $53,987.
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A quantity and its 2/3 are added together and from thesum 1/3 of the sum is subtracted, and 10 remains.What is the quantity?
Answer:
[tex]14\frac{1}{3}[/tex]
Step-by-step explanation:
Let's write this out as an equation. Let the unknown quantity be x:
[tex]x+\frac{2}{3} - \frac{1}{3} (x+\frac{2}{3}) = 10\\\\3x+2-x -\frac{2}{3} = 30\\\\9x+6-3x-2=90\\\\6x= 86\\\\x = \frac{86}{6} =\frac{43}{3} =14\frac{1}{3} \\[/tex]
Answer: the quantity is 9
Step-by-step explanation:
Let x represent the quantity.
A quantity and its 2/3 are added together. The 2/3 of the number is 2/3 × x = 2x/3
The sum of the quantity and its 2/3 would be
x + 2x/3 = (3x + 2x)/3 = 5x/3
From the sum, 1/3 of the sum is subtracted. 1/3 of the sum would be
1/3 × 5x/3 = 5x/9
Subtracting 1/3 of the sum from the sum, it becomes
5x/3 - 5x/9 = (15x - 5x)/9 = 10x/9
If the remainder is 10, it means that
10x/9 = 10
Crossmultiplying
10x = 9 × 10 = 90
x = 90/10
x = 9
in the function fx=a(x+2)(x-3)^b, a and b are bothj integer constants and b is positive, find the value of x for f(x)=0
Answer:
f(x) = 0 , when x=(-2) and x=3
Step-by-step explanation:
for a≠0 and b positive then for the function
f(x) = a*(x+2)*(x-3)ᵇ
f(x) = 0 when
1) (x+2)=0 → x=(-2)
2) (x-3)=0 → x=-3
then for x=3 and x=(-2) , f(x)=0
Find the matrix P that multiplies (x; y; z) to give (y; z; x). Find the matrix Q that multiplies (y; z; x) to bring back (x; y; z).
Answer:
[tex]P = \left[\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right][/tex]
[tex]Q = \left[\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right][/tex]
Step-by-step explanation:
[tex]P*\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}y\\z\\x\end{array}\right][/tex]
[tex]Q*\left[\begin{array}{c}y\\z\\x\end{array}\right]=\left[\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right] \left[\begin{array}{c}y\\z\\x\end{array}\right] = \left[\begin{array}{c}x\\y\\z\end{array}\right][/tex]
Determine if the statement is true or false. A system in echelon form can have more variables than equations. True False Justify your answer.
Answer:
Yes
Step-by-step explanation:
A system in echelon form can have more variables than equations. ... Every linear system with free variables has infinitely many solutions. True, a free variable can take any value, and so there are infinitely many solutions. Any linear system with more variables than equations cannot have a unique solution.
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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Convert 9255 cm3 to gallons. Express the volume in gallons to four significant figures.
Answer:
2.44491234
Step-by-step explanation:
1 gallon = 3785 cm3
Unit conversion is a way of converting some common units into another without changing their real value. The volume of 9255 cm³ in gallons is equal to 2.4451 gallons.
What is Unit conversion?Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.
Since 1 gallon is equal to 4546.09 cubic centimeters, therefore, the volume of 9255 cm³ in gallons can be written as,
1 gallons = 3785 cm³
1 cm³ = 1/3785 gallons
The volume of 9255 cm³ in gallons is,
Volume = 9255 × (1/3785 ) gallons
= 2.4451 gallons
Hence, the volume of 9255 cm³ in gallons is equal to 2.4451 gallons.
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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.
Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than 1.15 is:_________
Answer:
0.8749
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 = 0, \sigma = 1[/tex]
The probability that Z is less than 1.15 is:
This is the pvalue of Z = 1.15, which is 0.8749.
The probability that Z is less than 1.15: P(Z < 1.15) = Φ(1.15) ≈ 0.8749 or 87.49%
For a standard normal distribution with a mean of 0 and a standard deviation of 1, the probability that the random variable Z is less than 1.15 can be calculated using the standard normal cumulative distribution function (CDF).
The standard normal CDF gives the probability that a standard normal random variable takes a value less than or equal to a given value. We can use a standard normal distribution table or a calculator/software to find the value of the CDF at the desired point.
Let's denote the CDF of the standard normal distribution as Φ(z). Then, the probability that Z is less than 1.15 is given by:
P(Z < 1.15) = Φ(1.15)
Using a standard normal distribution table or calculator, we find that Φ(1.15) ≈ 0.8749.
Therefore, the probability that Z is less than 1.15 is approximately 0.8749 or 87.49%.
PLEASE HELP! See attached image for the equation!
Answer:
Step-by-step explanation:
The first step is to rearrange the given quadratic equation so that it will take the form of the standard quadratic equation which is expressed as
ax² + bx + c = 0
The given quadratic equation is expressed as
2 = - x + x² - 4
Rearranging it, we would subtract 2 from the left hand side and the right hand side of the equation. It becomes
2 - 2 = - x + x² - 4 - 2
- x + x² - 6 = 0
x² - x - 6
a = 1
b = - 1
c = - 6
Substituting into the quadratic formula, it becomes
x = [- (- 1) ± √(- 1)² - 4(1)(- 6)] /2(1)
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 11,12,13,14,15A 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.
Can some body pls help me
Answer:
Step-by-step explanation:
The diagram consists of different boxes if the same sizes. Each box is a square with sides measuring 1 cm. The area of each box would be
1 × 1 = 1cm^2
The area of the parallelogram is the total number of squares that it contains. To determine this, the first step is to count the total number of complete squares. Looking at the diagram, there are 11 complete squares. This means
11 × 1 = 11 square units.
The next step is to count the number of incomplete squares and divide by 2. The number of incomplete squares are 8
8/2 = 4
4 × 1 = 4 square units
The area of the parallelogram would be
11 + 4 = 15 square units
the absolute value of a complex number z=a+bi, denoted |z|, is given by the formula
|z|=square root a^2+b^2.
find six distinct complex numbers (which are neither purely imaginary nor purley real) such that each has an absolute value of 7.
Step-by-step explanation:
You need to find six combinations of a and b such that a² + b² = 7.
a = 1, b = √6
a = √2, b = √5
a = √3, b = 2
a = 2, b = √3
a = √5, b = √2
a = √6, b = 1
A person walks in the following pattern: 2.4 km north, then 1.9 km west, and finally 4.7 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?
Answer:
a) Figure attached
b) For this case w have that A =(2.4 km)j, B= (-1.9 km) i , C= (-4.7 km)j
And the final position vector can be calculated adding the 3 vectors like this:
[tex] s = A +B+C[/tex]
[tex] s= (-1.9 km)i +(2.4 -4.7 km) j= (-1.9km)i + (-2.3 km)j[/tex]
We can find the magnitude of s like this:
[tex] |s| = \sqrt{(-1.9)^2 +(-2.3)^2}=2.983[/tex]
And then we can find the angle with this formula:
[tex] \theta = \tan^{-1} (\frac{-2.3 km}{-1.9 km})=50.44 [/tex]
The other possibility is [tex] \theta = 50.44+180 =230.44[/tex]
And since they want the angle measured from East the correct angle would be [tex] \theta = 230.44[/tex]
Step-by-step explanation:
Part a
On the figure attached we have the vectors for the pattern described.
Part b
For this case w have that A =(2.4 km)j, B= (-1.9 km) i , C= (-4.7 km)j
And the final position vector can be calculated adding the 3 vectors like this:
[tex] s = A +B+C[/tex]
[tex] s= (-1.9 km)i +(2.4 -4.7 km) j= (-1.9km)i + (-2.3 km)j[/tex]
We can find the magnitude of s like this:
[tex] |s| = \sqrt{(-1.9)^2 +(-2.3)^2}=2.983[/tex]
And then we can find the angle with this formula:
[tex] \theta = \tan^{-1} (\frac{-2.3 km}{-1.9 km})=50.44 [/tex]
The other possibility is [tex] \theta = 50.44+180 =230.44[/tex]
And since they want the angle measured from East the correct angle would be [tex] \theta = 230.44[/tex]
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%.
Learn more about Probability here:https://brainly.com/question/22962752
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Which of the following equations is also true?
Answer:
cos 38 = 17/c
Step-by-step explanation:
cos 38 = 17/c is the correct statement
Suppose the length of the vegetable garden is increased by 50%. The new length can be written as (3y + 5) + 1 2(3y + 5) or 1.5(3y + 5). Explain why both expressions are correct. Then write the new length of the garden in simplified form.
Answer:
Given:
Length of the garden is represented by :
[tex]3y+5[/tex]
The length is increased by 50%.
To find the new length.
Solution:
1) The original length = [tex]3y+5[/tex]
The increase in length will be = [tex]50%[/tex] of [tex]3y+5[/tex]
⇒ [tex]\frac{50}{100}(3y+5)[/tex]
⇒ [tex]\frac{1}{2}(3y+5)[/tex]
The new length = Original length + Increased length = [tex](3y+5)+ \frac{1}{2}(3y+5)[/tex]
2) Let the original length = 100%
Increase percent of length = 50%
Total percent of length = [tex]100\%+50\%=150\%[/tex]
New length = [tex]150%[/tex] of [tex]3y+5[/tex]
⇒ [tex]\frac{150}{100}(3y+5)[/tex]
⇒ [tex]1.5(3y+5)[/tex]
Thus, the new length of the garden can be represented by both expression :
[tex](3y+5)+ \frac{1}{2}(3y+5)[/tex] or [tex]1.5(3y+5)[/tex]
Suppose x x represents the value of some varying quantity. As the value of x x varies from 3.5 to 8.
What is the change in the value of x x?
Answer: The change in the value of x would be : [tex]\Delta x=4.5 [/tex] .
Step-by-step explanation:
We know that if y represents the value of some varying quantity, then as the value of y varies from p to q.
The change in the value of y is given by :-
[tex]\Delta y=q-p[/tex] [Next value - previous value]
Given : x represents the value of some varying quantity.
So , As the value of x varies from 3.5 to 8.
The change in the value of x is given by :-
[tex]\Delta x=8-3.5=4.5 [/tex]
Hence, the change in the value of x would be : [tex]\Delta x=4.5 [/tex]
Observations on burst strength (Ib/in2) were obtained both for test nozzle closure welds and for production canister nozzle welds. Test 7200 6100 7300 7300 8000 7400 7300 7300 8000 6700 8300 5250 5625 5900 5900 5700 6050 5800 6000 5875 6100 5850 6600 Cannister A comparative boxplot is given below. Test Nozde Cannister 5000 000006000 Bunt strength (b, m Comment on interesting features (the cited article did not include such a picture, but the authors commented that they had looked at one). (Select all that apply.) The production canister welds have consistently lower burst strengths than the test nozze welds. The production canister weids have much more variable burst strengths The test nozzle welds data contain 2 outliers. test nozzle welds have much more variable burst strengths. The production canister welds have much higher burst strengths. The production canister welds data contain 2 outiers.
Answer:
1) The production canister welds have consistently lower burst strengths than the test nozze welds.
2) The production canister weids have much more variable burst strengths.
3) The test nozzle welds data contain 2 outliers.
4) Test nozzle welds have much more variable burst strengths.
5) The production canister welds have much higher burst strengths.
6) The production canister welds data contain 2 outiers.
Step-by-step explanation:
Hello!
The boxplots summarize the information of test nozzle closure welds and production canister nozzle welds.
The boxplot for the test nozzle closure welds shows that the first quartile and second quartile are close to each other but the third quartile is more separated to them, meaning that the data contained in the box is asymmetric, the data seems to have less variability between C₁ and C₂ and more between C₂ and C₃, the box is right-skewed.
The left whisker is larger than the right one, there are no outliers in the sample, due to most of the data being comprehended below C₁, the overall distribution of the data set is left-skewed, with large variability.
The boxplot for production cannister nozzle welds shows that the box is small (the variability of the data set is low) and symmetric, with C₂ in the middle of it and C₁ and C₃ are equidistant to the second quartile.
The whiskers of the box are small but they have almost the same length, showing that there is the same amount of data in them, this adds to the overall symmetry of the data set.
Finally, this data set shows two outliers, these values are far from the box, meaning that they are relatively extreme unusual values in regards to the rest of the sample but their distance to the box seems to be equal wich adds to the conclusion of the symmetrical distribution, with low variability of the data set.
I hope it helps!