Walmart's decision to use its own fleet of trucks to pick up products directly from manufacturers and deliver them to its stores is a strategic move that has the potential to benefit both Walmart and manufacturers.
By using its own fleet, Walmart will be able to cut down on transportation costs and gain more control over the supply chain, which can lead to better efficiency and cost savings. This is especially important given Walmart's massive scale, with over 4,000 stores in the US alone.
For manufacturers, this move by Walmart could be a relief as they can focus on their core competency of making products rather than managing logistics. With Walmart taking over the transportation aspect, manufacturers can rest assured that their products will be delivered on time and in the right condition.
The fact that Walmart already has a large fleet of trucks and trailers means that it has the capacity to implement this new program without too much additional investment. However, it remains to be seen how manufacturers will respond to this proposal, as they may have existing contracts with other carriers or may be hesitant to rely too heavily on Walmart for their transportation needs.
Overall, Walmart's move towards using its own fleet of trucks is a smart one that has the potential to benefit both the retailer and its suppliers.
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Find The Linear Approximation To The Function F (Dy, Z) = Ce2yz+32 At The Point (X, Y, Z) = (3,-2,0)
f(x,y,z) = xe^2yz+3z
L (x,y,z) =
At the coordinates (X, Y, Z) = (3, -2, 0), L(x, y, z) = C+35 + x - 12Cz is the linear approximation to the function f(Dy, Z) = Ce^2yz+32.
To find the linear approximation to the function f(Dy, Z) = Ce^2yz+32 at the point (X, Y, Z) = (3,-2,0), we need to find the partial derivatives of the function with respect to each variable at the point (3,-2,0).
The partial derivative of f with respect to x is simply e^2yz, which evaluated at (3,-2,0) gives us e^0 = 1.
The partial derivative of f with respect to y is 2xzCe^2yz, which evaluated at (3,-2,0) gives us 2(3)(0)C = 0.
The partial derivative of f with respect to z is 2xyCe^2yz+3, which evaluated at (3,-2,0) gives us 2(3)(-2)C + 3(1) = -12C + 3.
Using these partial derivatives, we can construct the linear approximation L(x,y,z) = f(3,-2,0) + (x-3)(1) + (y+2)(0) + (z-0)(-12C+3) = Ce^0+32 + (x-3) - 12Cz + 3.
Simplifying this expression, we get L(x,y,z) = C+35 + x - 12Cz.
Therefore, the linear approximation to the function f(Dy, Z) = Ce^2yz+32 at the point (X, Y, Z) = (3,-2,0) is L(x,y,z) = C+35 + x - 12Cz.
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What in 33/22 x 44/33 equal?
Answer:
Step-by-step explanation:
which of the following statements about a randomly chosen person from these 200 employees is true? responses if the person owns a car, he or she is more likely to live elsewhere in the city than to live in the downtown area in the city. if the person owns a car, he or she is more likely to live elsewhere in the city than to live in the downtown area in the city. if the person does not own a car, he or she is more likely to live outside the city than to live in the city (downtown area or elsewhere). if the person does not own a car, he or she is more likely to live outside the city than to live in the city (downtown area or elsewhere). the person is more likely to own a car if he or she lives in the city (downtown area or elsewhere) than if he or she lives outside the city. the person is more likely to own a car if he or she lives in the city (downtown area or elsewhere) than if he or she lives outside the city. the person is more likely to live in the downtown area in the city than elsewhere in the city. the person is more likely to live in the downtown area in the city than elsewhere in the city. the person is more likely to own a car than not to own a car.
The statement that if a person has his own car, then there is more chances that he or she live elsewhere in the city than to live in the downtown area in the city is true statement. So, the option(a) is right answer for problem.
We have, a sample of sample size, n
= 200
The above table contains data about location of home ( that is downtown area in city, elsewhere in city, outside the city and total) and car ownership ( Yes or No ). Now, we determine probability that car ownership |downtown area in the city
= 10/70
= 1/7
probability that car ownership | elsewhere in the city = 15/70
= 3/14
Probability that car ownership | outside in the city = 35/60 = 7/12
As we see, 1/7 < 3/14 < 7/12
here probability of car ownership downtown area of city is less than probability of car ownership if lives elsewhere in the city or less than probability of car ownership if outside the city. So, statement (a) is true in this case. Also consider,
Probability that no car ownership | downtown area in the city = 60/70 = 6/7
Probability that no car ownership | elsewhere in the city = 55/70 = 11/14
Probability that no car ownership|outside the city = 25/60 = 5/12
As we see probabilities, 5/12 < 11/14 < 6/7
Here, if a person has not own car then maximum chances that he or she live in downtown area of city. So, statement (b) is false.
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Complete question:
The above figure complete the question. A local company is interested in supporting environmentally friendly initiatives such as carpooling among employees. The company surveyed all of the 200 employees at the downtown offices. Employees responded as to whether or not they own a car and to the location of the home where they live. The results are shown in the table above. Which of the following statements about a randomly chosen person from these 200 employees is true? responses
a) if the person owns a car, he or she is more likely to live elsewhere in the city than to live in the downtown area in the city.
b) if the person does not own a car, he or she is more likely to live outside the city than to live in the city (downtown area or elsewhere).
c) the person is more likely to own a car if he or she lives in the city (downtown area or elsewhere) than if he or she lives outside the city.
d) the person is more likely to live in the downtown area in the city than elsewhere in the city.
e) the person is more likely to own a car than not to own a car.
All of the training times of which person had the greatest spread? Explain how you know. (b) The middle 50% of the training times of which person had the least spread? Explain how you know. (c) What do the answers to Parts 2(a) and 2(b) tell you about Adam’s and Miguel’s training times?
(a) Miguel had the greatest spread in training times.
(b) The middle 50% of Adam's training times had the least spread.
(a) To find the greatest spread in training times, we need to calculate the range of each person's training times. Range is the difference between the maximum and minimum values. Comparing the ranges, we can say that Miguel had the greatest spread in training times since his range is the largest.
(b) The middle 50% of the training times refers to the interquartile range (IQR), which is the difference between the third quartile (Q3) and the first quartile (Q1).
To find the least spread in the middle 50% of the training times, we need to compare the IQRs of each person. Adam's IQR is the smallest, which means the middle 50% of his training times had the least spread.
(c) The answers to parts (a) and (b) indicate that Miguel had a wider range of training times compared to Adam. However, Adam's middle 50% of training times had the least spread. This suggests that while Miguel's overall training times varied more, Adam's training times were more consistent within the middle range.
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Determine the intercepts of the line.
Do not round your answers.
-5x - 4y = 10
Step-by-step explanation:
-5x - 4y = 10
Intercept-y (x = 0)
-5 (0) - 4y = 10
-4y = 10
y = - 5/2
(0, -5/2)
Intercept-x (y = 0)
-5x - 4 (0) = 10
-5x = 10
x = -2
(-2, 0)
#CMIIWpoint P is the image of p(-2,-2) translated by 1 unit to the left and 3 units now
To find the image of point P(-2,-2) translated 1 unit to the left and 3 units down, we subtract 1 from the x-coordinate and 3 from the y-coordinate to get coordinates of P' as: (-3, -5).
How to Find the Coordinates in Translation?To translate a point to the left, we subtract from its x-coordinate, and to translate it down, we subtract from its y-coordinate.
Therefore, to translate P(-2, -2) 1 unit to the left and 3 units down, we subtract 1 from the x-coordinate and 3 from the y-coordinate to get P'(-3, -5) as the image of P after translation.
Therefore, the coordinates of P' are (-3, -5).
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Complete Question:
Point P' is the image of P(-2,-2) translated by 1 unit to the left and 3 units down. What are the coordinates of P'?
The percentage of the moon's surface that is visible to a person standing on the Earth varies with the time
since the moon was full.
The moon passes through a full eyele in 28 days, from full moon to full moon. The
maximum percentage of the moon's surface that is visible is 50%. Determine an equation, in the form
P=Acos(Bt)+C for the percentage of the surface that is visible, P, as a function of the number of days, t,
since the moon was full. Show the work that leads to the values of A, B, and C
The equation is P = [tex]25cos(0.224t) + 50[/tex], where P represents the percentage of the moon's surface visible and t is the number of days since the moon was full.
How to derive equation for moon visibility?To determine an equation for the percentage of the moon's surface visible as a function of the number of days since the moon was full, we can use the cosine function [tex]P = Acos(Bt) + C[/tex], where P represents the percentage visible, t is the number of days since full moon, A is the amplitude, B is the frequency, and C is the vertical shift.
Given that the maximum percentage visible is 50%, we know that C = 50. The period of the function is 28 days, so we can calculate B using the formula B = 2π/period = 0.224. The amplitude A can be calculated as half of the maximum percentage visible, or A = 25.
Therefore, the equation for the percentage of the moon's surface visible as a function of the number of days since full moon is P = 25cos(0.224t) + 50.
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The mean number of sit-ups
done by a group of students is
46 with a standard deviation
of 7. If Rylee's Z-score was
1. 8, how many sit ups did she
do?
Rylee did approximately 58.6 sit-ups.
We are given that the mean number of sit-ups is 46 and the standard deviation is 7. We are also given that Rylee's Z-score was 1.8, we can use the formula for Z-score to find how many sit-ups she did.
The formula for Z-score is [tex]Z = \frac{X-\mu}{\sigma}[/tex]
Z = Z-score
μ = mean
σ = standard deviation
X = ?
Substituting these values into the formula
1.8 = (X - 46)/7
1.8 × 7 = X - 46
X - 46 = 12.6
X = 12.6 + 46
X = 58.6
Therefore, Rylee did approximately 58.6 sit-ups.
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Ann selects a sample of 29 students at her large high school and finds that 12 of them are planning to travel outside of the state during the coming summer. She wants to construct a confidence interval for p = the proportion of all students at her school who plan on traveling outside of the state during the coming summer, but she realizes she hasn’t met all the conditions for constructing the interval. Which condition for this procedure has she failed to meet?
Ann has failed to meet the condition called the "success-failure" condition.
In order to construct a confidence interval for the proportion (p), the sample must have at least 10 successes (planning to travel outside the state) and 10 failures (not planning to travel outside the state). In her sample of 29 students, she found 12 planning to travel (successes) and 17 not planning to travel (failures). Both numbers satisfy the success-failure condition, so she can construct the confidence interval for the proportion of students planning to travel outside the state during the coming summer.
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Answer:
C: The sample must be a random sample from the population
Step-by-step explanation:
took the test on edge
Divide.
Simplify your answer as much as possible.
Answer:
-5z/v³ + 8 + 6v²z
Do you have to simplify it further or?
Step-by-step explanation:
[tex] \frac{ - 5z + 8v {}^{3} + 6v {}^{5} z}{v {}^{3} } = \frac{ - 5z}{v {}^{3} } + \frac{8v {}^{3} }{ {v}^{3} } + \frac{6v {}^{5} }{v {}^{3} } = \frac{ - 5z}{v {}^{3} } + 8 + 6v {}^{2}z[/tex]
Differentiate. f(x)= In (x⁸-2/x) Differentiate. y =In (9x²-7x+4)
The derivative of y = ln[tex](9x^2-7x+4)[/tex] is y' = (18x-7) / [tex](9x^2-7x+4)[/tex].
To differentiate f(x) = ln[tex]((x^8-2)/x[/tex]), we use the chain rule and the quotient rule:
f'(x) = [[tex](x^8[/tex]-2)/x]' / (x^8-2)/x + ln[tex]((x^8-2[/tex])/x)'
[tex]= [((x^8-2)'x - (x^8-2)x') / x^2] / (x^8-2)/x + [(1/x)'(x^8-2) - (1)'x(x^8-2)/x^2][/tex]
[tex]= [(8x^7)(x) - (x^8-2)] / x^2(x^8-2)/x + [(1/x)(x^8-2)/x^2] - (1)(x^8-2)/x^2[/tex]
[tex]= [(8x^8-2-x^8+2)] / x(x^8-2) + [(x^8-2)/x^2(-x)][/tex]
[tex]= (7x^8-4) / (x^2(x^8-2)) - (x^8-2) / (x^3(x^8-2))[/tex]
Simplify to get:
[tex]f'(x) = (6x^8-4) / (x^3(x^8-2))[/tex]
Therefore, the derivative of f(x) = ln[tex]((x^8-2)/x) is f'(x) = (6x^8-4) / (x^3(x^8-2)).[/tex]
To differentiate y = ln[tex](9x^2-7x+4)[/tex], we use the chain rule:
y' =[tex][(9x^2-7x+4)' / (9x^2-7x+4)][/tex]
[tex]= [(18x-7) / (9x^2-7x+4)][/tex]
Simplify to get:
[tex]y' = (18x-7) / (9x^2-7x+4)[/tex]
Therefore, the derivative of y = [tex]ln(9x^2-7x+4) is y' = (18x-7) / (9x^2-7x+4).[/tex]
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In ΔEFG, g = 5. 2 cm, e = 5. 1 cm and ∠F=42°. Find the area of ΔEFG, to the nearest 10th of a square centimeter
The area of ΔEFG is approximately 6.7 square centimeters.
To find the area of ΔEFG with given sides g = 5.2 cm, e = 5.1 cm, and ∠F = 42°, you can use the formula for the area of a triangle when two sides and the included angle are known. This formula is:
Area = (1/2)ab * sin(C)
In this case, a = g, b = e, and C = ∠F. Plug in the values:
Area = (1/2)(5.2 cm)(5.1 cm) * sin(42°)
Area ≈ 6.675 square centimeters
So, the area of ΔEFG is approximately 6.7 square centimeters to the nearest 10th of a square centimeter.
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Let $f(x)=3x+2$ and $g(x)=ax+b$, for some constants $a$ and $b$. If $ab=20$ and $f(g(x))=g(f(x))$ for $x=0,1,2\ldots 9$, find the sum of all possible values of $a$
The sum of all possible values of $a$ is $1$.
To solve this problem, we need to use the given information to determine possible values of $a$ and $b$ in $g(x)=ax+b$ such that $f(g(x))=g(f(x))$ for $x=0,1,2\ldots 9$.
First, we can simplify $f(g(x))$ and $g(f(x))$ as follows:
$$f(g(x))=3(ax+b)+2=3ax+3b+2$$
$$g(f(x))=a(3x+2)+b=3ax+ab+b$$
Next, we can set these two expressions equal to each other and simplify:
$$3ax+3b+2=3ax+ab+b$$
$$2b-ab=b$$
$$(2-a)b=b$$
Since $ab=20$, we have two cases to consider:
Case 1: $b=0$
In this case, we have $ab=20\implies a=0$ or $b=0$. Since we are looking for non-zero values of $a$, we can eliminate $a=0$ and conclude that $b=0$. However, $b=0$ does not satisfy the given equation $f(g(x))=g(f(x))$, so there are no solutions in this case.
Case 2: $b\neq 0$
In this case, we can divide both sides of $(2-a)b=b$ by $b$ to get:
$$2-a=1$$
$$a=1$$
Therefore, the only possible value of $a$ is $1$, and the corresponding value of $b$ is $20$. We can verify that $a=1$ and $b=20$ satisfy the given equation $f(g(x))=g(f(x))$ for $x=0,1,2\ldots 9$.
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1. The following data show weight (in kg) of 24 women in a study: 46. 4, 53. 2, 52. 8, 42. 0, 50. 8,
43. 0, 51. 9, 59. 2, 55. 1, 38. 9, 49. 7, 49. 9, 43. 1,42. 2, 52. 7. 49. 8. 50. 7, 44. 8. 49. 2, 47. 7, 42. 9,
52. 9, 54. 1, 45. 4.
Prepare the following:
I.
Calculate a) mean, b) median, c) mode, d) variance, e) standard deviation, f)
coefficient variation, g) IQR
Box and whisker plot
II.
III.
Discuss the distribution of these data
The mean is 48.47 kg, median is 49.55 kg, mode is not available, variance is 34.1 kg², standard deviation is 5.84 kg, coefficient of variation is 12.03% and IQR is 8.35 kg.
The given data shows the weight (in kg) of 24 women in a study. To analyze the data, we need to calculate various statistical measures:
I. Statistical Measures:
a) Mean = (Sum of all weights) / (Number of observations) = (1163.4) / (24) = 48.47 kg
b) Median = Middle value of the sorted data set = 49.55 kg
c) Mode = The most frequent value in the data set = No mode as there are no repeating values.
d) Variance = (Sum of squares of deviations of each value from mean) / (Number of observations) = 34.1 kg²
e) Standard deviation = Square root of variance = 5.84 kg
f) Coefficient of variation = (Standard deviation / Mean) x 100 = 12.03%
g) IQR (Interquartile range) = Q3 - Q1 = 53.025 - 44.675 = 8.35 kg
II. Box and Whisker Plot:
The box and whisker plot displays the distribution of the data. The lower and upper quartiles are represented by the bottom and top of the box respectively, and the median is represented by the line in the middle. The whiskers represent the minimum and maximum values.
III. Distribution:
The data set appears to be skewed to the right as the median is less than the mean. There are no outliers in the data, and the IQR is relatively small, indicating that the data is not too spread out. The coefficient of variation is moderate, indicating that the data has a moderate degree of variation. Overall, the data set seems to be fairly normal, with a few outliers on the right side.
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Evaluate f(x) = 7x2 − 8 when x = 5.
Answer:
f(5) = 167
Step-by-step explanation:
To evaluate f(x) = 7x^2 - 8 when x = 5, we substitute 5 for x in the expression and simplify. Therefore, we have:
f(5) = 7(5)^2 - 8
f(5) = 7(25) - 8
f(5) = 175 - 8
f(5) = 167
So, f(5) = 167
A ship headed due east is moving through the water at a constant speed of 8 miles per hour. However, the true course of the ship is 60°. If the currents are a constant 4 miles per hour, what is the ground speed of the ship? (Round your answer to the nearest whole number. )
The ground speed of the ship is approximately 10 miles per hour.
To calculate the ground speed, we need to use vector addition. The ship's velocity can be broken down into two components: its speed in the easterly direction and its speed in the northerly direction. The easterly component is 8 miles per hour (since the ship is moving due east), and the northerly component can be found using trigonometry: northerly component = 8 * sin(60°) ≈ 6.93 miles per hour
Now, we need to take into account the effect of the currents, which are moving in a southerly direction. Again using vector addition, we can find the resultant velocity (i.e., the velocity of the ship relative to the ground) by adding the ship's velocity vector to the current's velocity vector. Since the current is moving due south, its velocity vector has no easterly component, but its southerly component is 4 miles per hour. resultant velocity = (8, 6.93) + (0, -4) = (8, 2.93)
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity: |resultant velocity| = [tex]\sqrt{} (8^2 + 2.93^2)[/tex]≈ 8.6 miles per hour. Rounding to the nearest whole number, the ground speed of the ship is approximately 10 miles per hour.
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In 2012, the population of a city was 6.47 million. the exponential growth rate was 2.91% per year.
The population of the city after 5 years is approximately 7.94 million.
How to find the population of the city?Assuming that the population of the city grows exponentially, we can use the formula:
P(t) = [tex]P0 * e^(^r^t^)[/tex]
Where:
- P(t) is the population after time t
- P0 is the initial population
- r is the annual growth rate expressed as a decimal
- t is the time elapsed in years
Using the given information:
- P0 = 6.47 million
- r = 2.91% = 0.0291
Let's calculate the population after 1 year:
[tex]P(1) = 6.47 million * e^(^0^.^0^2^9^1 ^* ^1^)[/tex]
= 6.66 million (rounded to two decimal places)
So, the population of the city after one year is approximately 6.66 million.
We can also calculate the population after 5 years:
[tex]P(5) = 6.47 million * e^(^0^.^0^2^9^1 ^* ^5^)[/tex]
= 7.94 million (rounded to two decimal places)
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Write exponential functions given the following scenarios:
1. a business had a profit of $35,000 in 1998 that increased by 18% per year. write the equation to model the
situation. find the profit of the company after 8 years.
2. you buy a used truck for $4,000. the value of the truck depreciates at a yearly rate of 12%. write the equation to model the situation. find the value of the truck after 6 months.
3. between 1970 and 2000, the population of a town increased by approximately 2.5% each year. in 1970 there were 600 people. write the equation to model the situation. find the population of the city in 1999.
The profit of the company after 8 years is approximately $105,085.11.
The value of the truck after 6 months is approximately $3,677.49.
The population of the city in 1999 is approximately 1,457.66 people.
How we write the exponential functions?Let P(t) be the profit in year t, where t is the number of years after 1998. The initial profit in 1998 is $35,000.
The profit increases by 18% per year, which means the profit at time t is 1.18 times the profit at time t-1. Therefore, the equation to model the situation is: [tex]P(t) = 35000 * 1.18^t[/tex]
To find the profit of the company after 8 years:
[tex]P(8) = 35000 * 1.18^8[/tex] = $105,085.11
Let V(t) be the value of the truck in year t, where t is the number of years after the purchase. The initial value of the truck is $4,000.
The value depreciates at a yearly rate of 12%, which means the value at time t is 0.88 times the value at time t-1. Therefore, the equation to model the situation is: [tex]V(t) = 4000 * 0.88^t[/tex]
To find the value of the truck after 6 months (0.5 years):
[tex]V(0.5) = 4000 * 0.88^0^.^5[/tex] = $3,677.49
Let P(t) be the population of the town in year t, where t is the number of years after 1970. The initial population in 1970 is 600.
The population increases by 2.5% per year, which means the population at time t is 1.025 times the population at time t-1. Therefore, the equation to model the situation is: [tex]P(t) = 600 * 1.025^t[/tex]
To find the population of the city in 1999 (29 years after 1970):
[tex]P(29) = 600 * 1.025^2^9 = 1,457.66[/tex]
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Which table has a constant of proportionality between
�
yy and
�
xx of
12
1212?
Choose 1 answer:
Choose 1 answer:
(Choice A)
�
xx
�
yy
1
2
2
1
start fraction, 1, divided by, 2, end fraction
6
66
2
22
24
2424
10
1010
120
120120
A
�
xx
�
yy
1
2
2
1
start fraction, 1, divided by, 2, end fraction
6
66
2
22
24
2424
10
1010
120
120120
(Choice B)
�
xx
�
yy
1
4
4
1
start fraction, 1, divided by, 4, end fraction
3
33
3
33
60
6060
12
1212
144
144144
B
�
xx
�
yy
1
4
4
1
start fraction, 1, divided by, 4, end fraction
3
33
3
33
60
6060
12
1212
144
144144
(Choice C)
�
xx
�
yy
1
3
3
1
start fraction, 1, divided by, 3, end fraction
4
44
6
66
78
7878
9
99
117
117117
C
�
xx
�
yy
1
3
3
1
start fraction, 1, divided by, 3, end fraction
4
44
6
66
78
7878
9
99
117
117117
The table that have a constant of proportionality between y and x of 12 is the first table
What is the table that have a constant of proportionality between y and x of 12?From the question, we have the following parameters that can be used in our computation:
The table of values
From the first table of values, we have the following readings
(x, y) = (1/2, 6), (2, 24) and (10, 120)
Using the above as a guide, we have the following:
The constant of proportionality between y and x in the graph is
k = y/x
Substitute the known values in the above equation, so, we have the following representation
k = 6/(1/2) = 24/2 = 120/10
Evaluate
k = 12 = 12 = 12
Hence, the constant of proportionality between y and x in the first table is 12
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Complete question
Which table has a constant of proportionality between y and x of 12?
x 1/2 2 10
y 6 24 120
x 1/4 3 12
y 3 60 144
x 1/3 6 9
y 4 78 117
what does 8 thousands plus 8 tens equal?
Answer:
100
Step-by-step explanation:
8000/80=100
8 thousands= 8000
8 tens= 80
Expert Answer, Mark AS BRAINLIEST!!!
Four buses carrying 150 football fans from the same school arrive at a football stadium. The buses carry, respectively, 20, 45, 35, and 50 students. One of the fans is randomly selected. Let X denote the number of fans that were on the bus carrying the randomly selected person. One of the 4 bus drivers is also randomly selected. Let Y denote the fans of students on his bus. Compute E(X) and Var(X)
If 4 buses carrying 150 football fans from same school arrive at a football stadium, then the expected-value, "E(X)" is 41 and variance "Var(X)" is 99.
To find the expected value of X, we use the formula E(X) = ∑x P(X=x), where x is = possible values of X and P(X=x) = probability of X taking the value x.
Four buses have a total of 150 students, the probability that the randomly selected person is from a bus with x students is the proportion of students on that bus divided by the total number of students:
P(X=x) = (number of students on bus with x students)/(total number of students);
So, We have:
P(X=20) = 20/150 = 2/15
P(X=35) = 35/150 = 7/30
P(X=45) = 45/150 = 3/10
P(X=50) = 50/150 = 1/3
The expected-value of X is : E(X) = 20(2/15) + 35(7/30) + 45(3/10) + 50(1/3) = 41
To find the variance of X, we use the formula Var(X) = E(X²) - [E(X)]².
We already know E(X), so we need to find E(X²).
E(X²) = ∑ x² P(X=x);
So, We have:
E(X²) = 20²(2/15) + 35²(7/30) + 45²(3/10) + 50²(1/3) = 1780;
So, variance of X is : Var(X) = E(X²) - [E(X)]² = 1780 - 41² = 99.
Therefore, the expected value of X is 41 and the variance of X is 99.
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pls help <3 Triangle QRS has side lengths q = 11, r = 17, and s = 23. What is the measure of angle R
a.44.5°
b.59.3°
c.27.0°
d.108.6
Using the cosine law, the measure of angle R is calculated as approximately: a. 44.5°.
How to Use the Cosine Law to Solve a Triangle?The cosine law is expressed as follows:
cos R = [s² + q² – r²]/2sq
Given the following side lengths of triangle QRS:
Side q = 11,
Side r = 17,
Side s = 23.
Plug in the values into the cosine law formula:
cos R = [23² + 11² – 17²]/2 * 23 * 11
cos R = 361/506
Cos R = 0.7134
R = cos^(-1)(0.7134)
R ≈ 44.5°
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A. Plot point C so that its distance from the origin is 1. B. Plot point E 4/5 closer to the origin than C. What is its coordinate? c. Plot a point at the midpoint of C and E. Label it H
(A). To plot point C so that its distance from the origin is 1, we need to find a point on the coordinate plane that is 1 unit away from the origin. One such point is (1, 0), which is located on the positive x-axis.
(B). To plot point E 4/5 closer to the origin than C, we need to find a point that is 4/5 of the distance from the origin to point C. Since point C is located 1 unit away from the origin, point E will be 4/5 of 1 unit away from the origin, or 0.8 units away.
To find the coordinates of point E, we can multiply the coordinates of point C by 0.8. If point C is (1, 0), then point E is (0.8, 0).
(C). To plot a point at the midpoint of C and E, we can use the midpoint formula, which is (x1 + x2)/2, (y1 + y2)/2.
The coordinates of point C are (1, 0) and the coordinates of point E are (0.8, 0), so the coordinates of point H are ((1 + 0.8)/2, (0 + 0)/2), or (0.9, 0). We can label this point H.
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Find the standard matrix for the linear transformation T:R2 + R2 that shears horizontally, with T "((A)) = (-1,67)
The standard matrix for the linear transformation T that shears horizontally is T = [(1 1) (0 1)] [(1 0) (-6 1)] [(1 1) (0 1)]^(-1) = [(1 -6) (0 1)].
To find the standard matrix for the linear transformation T that shears horizontally, we need to determine the matrix that transforms the standard basis vectors e1 and e2 into the shear vectors s1 and s2. The shear vectors are obtained by applying the linear transformation T to the standard basis vectors e1 and e2, respectively.
The shear vector s1 is obtained by shearing the point (1,0) horizontally by -1 unit, and then vertically by 6 units. This gives us s1 = (-1,6). Similarly, the shear vector s2 is obtained by shearing the point (0,1) horizontally by -1 unit and leaving it vertically unchanged. This gives us s2 = (-1,1).
To obtain the standard matrix for the linear transformation T, we need to find the matrix A that transforms the standard basis vectors e1 and e2 into the shear vectors s1 and s2, respectively. We can express A as [s1 s2] [e1 e2]^(-1), where [s1 s2] is a 2x2 matrix whose columns are the shear vectors, and [e1 e2]^(-1) is the inverse of the 2x2 matrix whose columns are the standard basis vectors.
Substituting the values of s1, s2, e1, and e2, we get:
A = [(1 -1) (6 1)] [(1 0) (0 1)]^(-1) = [(1 -1) (6 1)] [(1 0) (0 1)] = [(1 -1) (6 1)]
Therefore, the standard matrix for the linear transformation T that shears horizontally is T = [(1 1) (0 1)] [(1 -1) (6 1)] [(1 1) (0 1)]^(-1) = [(1 -6) (0 1)].
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Every day, Lucy's burrito stand uses 3/4 of a bag of tortillas. How many days will 3 3/4 bags of tortillas last?
The number of days 3 3/4 bags of tortillas will last is 5 days.
To solve this problem, we need to use the concept of fractions. We know that Lucy's burrito stand uses 3/4 of a bag of tortillas every day. So, if we want to find out how many days 3 3/4 bags of tortillas will last, we need to divide 3 3/4 by 3/4.
To do this, we can convert 3 3/4 to an improper fraction, which is 15/4. Then, we can divide 15/4 by 3/4 using the following steps:
15/4 ÷ 3/4 = 15/4 x 4/3 (we flip the second fraction and multiply)
= 60/12 (we simplify by finding a common denominator of 12)
= 5
Therefore, 3 3/4 bags of tortillas will last for 5 days at Lucy's burrito stand.
In conclusion, using fractions can help us solve real-life problems such as this one involving tortillas at a burrito stand. By understanding how to convert between mixed numbers and improper fractions, and how to divide fractions, we can calculate how long a given amount of tortillas will last and make informed decisions about our business operations.
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You are going to run at a constant speed of 7.5
miles per hour for 45
minutes. You calculate the distance you will run. What mistake did you make in your calculation? [Use the formula S=dt
.]
Answer: Did not convert the minutes to hours
Step-by-step explanation:
The most obvious mistake here would be to not convert the minutes into hours.
Remember, the speed given, (7.5), is in miles per HOUR. Your time is given in MINUTES. a conversion is required. 45 minutes are 45/60 = 0.75 hrs.
NOW you can use S = DT to find your distance:
7.5 = D/0.75
.: D = 5.625 miles
On a coordinate plane, triangle A B C has points (negative 2, 7), (negative 2, 3), and (negative 6, 3) and triangle D E F has points (negative 2, negative 10), (negative 2, negative 2), and (6, negative 2).
Given that StartFraction A B Over D E EndFraction = StartFraction B C Over E F EndFraction = one-half, complete the statements to show that △ABC ~ △DEF by the SAS similarity theorem.
Horizontal and vertical lines are
congruent
.
So, angles
are right angles by definition of perpendicular lines.
All right angles are
.
Therefore, △ABC ~ △DEF by the SAS similarity theorem.
please hurry A 4-column table with 3 rows. Column 1 has entries boys, girls, total. Column 2 is labeled less than 8 pounds with entries a, c, e. Column 3 is labeled greater-than-or-equal-to 8 pounds with entries 50, d, 70. Column 4 is labeled Total with entries b, 96, 160.
Last July, 160 babies were born in a hospital in Maine; 3
5
of the babies were girls. Seventy babies weighed 8 pounds or more. Fifty boys weighed 8 pounds or more.
a = 64, b = 14, c = 76, d = 20, e = 90
a = 14, b = 64, c = 90 d = 20, e = 76
a = 14, b = 76, c = 64, d = 90, e = 20
a = 14, b = 64, c = 76, d = 20, e = 90
Answer: the correct answer is:
a = 14, b = 64, c = 76, d = 20, e = 90
Step-by-step explanation: can i get brainliest :D
pa help, pleaseeee. sana mahanap to ng matalino, pumuputok utak ko, pahelp naman po please
Ms thompson sets up chairs in a row for a school concert. she uses 328. she sets up at 2 roses of chairs but not more than 10 rows of chairs each row has an equal number of chairs how many rows
Ms. Thompson could set up either 2 rows with 164 chairs in each row or 4 rows with 82 chairs in each row.
To find the number of chairs in each row, we need to divide the total number of chairs by the number of rows. Let's start by finding the factors of 328:
1 x 328
2 x 164
4 x 82
8 x 41
Since there must be at least 2 rows and no more than 10 rows, we can eliminate the last two factor pairs. We are left with:
2 x 164
4 x 82
We can see that the first factor pair gives us 2 rows, while the second gives us 4 rows. We are told that each row has an equal number of chairs, so we need to divide the total number of chairs by the number of rows to find out how many chairs are in each row:
For 2 rows: 328 ÷ 2 = 164 chairs in each row
For 4 rows: 328 ÷ 4 = 82 chairs in each row
Your question is incomplete but most probably your full question
Ms. Thompson sets up chairs in rows for a school concert. She uses 328 chairs. She sets up at least 2 rows of chairs but not more than 10 rows of chairs. Each row has an equal number of chairs.
How many rows of chairs does Ms. Thompson set up? Enter the number in the first box.
How many chairs are in each row? Enter the number in the second box.
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