Two people are trying to decide whether a die is fair. The correct test for analyzing the fairness of the die with the given data is the chi-square [tex]X^2[/tex] test, not the z-test.
The z-test is used for analyzing data when we have known population parameters, such as the mean and standard deviation. However, in this case, we are dealing with categorical data (the frequencies of each face of the die), and we want to determine if the observed frequencies significantly differ from the expected frequencies.
To perform the chi-square test, we first need to calculate the expected frequency for each face of the die. The expected frequency is calculated by multiplying the total number of rolls (100) by the probability of each face (1/6, assuming a fair die). Each face of the die is expected to occur approximately 16.67 times (100/6 = 16.67).
Next, we calculate the degrees of freedom (df) for the chi-square test. For a fair die with 6 faces, the df is (number of categories - 1), which is 5 in this case.
Then, we calculate the chi-square statistic[tex](X^2)[/tex] by summing the squared differences between the observed and expected frequencies, divided by the expected frequencies. The [tex]X^2[/tex] value is used to assess the goodness-of-fit between the observed and expected frequencies.
Finally, we determine the p-value associated with the calculated [tex]X^2[/tex]value using the chi-square distribution and the degrees of freedom. The p-value indicates the likelihood of observing the data if the die is fair.
To provide the specific values for the expected frequency, degrees of freedom, [tex]X^2[/tex], and p-value, the actual calculations based on the given data are required.
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Miguel wants to estimate the average price of a book at a bookstore. The bookstore has 13,000 titles, but Miguel only needs a sample of 200 books. How could Miguel collect a sample of books that is:
cluster sample?
multistage sample?
oversamples?
To best collect a sample of books from the 13,000 titles at the bookstore, Miguel should use a cluster sampling method.
What is the best sampling method?The best sampling method Miguel should use is a cluster sample.
A cluster sample involves dividing the population into clusters or groups and randomly selecting entire clusters to include in the sample.
In this situation, Miguel could divide the bookstore's titles into clusters, such as by genre or shelf location, and randomly select clusters to sample books from.
his method would help ensure representation from different areas of the bookstore and provide a diverse sample of books.
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Which graph represents an equation with the values shown in the table?
Answer:
4
Step-by-step explanation:
The table below gives a record of variations of the values of y with the values of x. Draw a scatter plot for the data.
x
0.4
1.2
2.0
3.1
4.5
5.7
7.1
8.4
9.3
9.8
y
7.8
7.1
6.8
6.0
5.2
4.3
3.4
2.3
1.1
0.5
a.
On a graph, points are at (2, 6.9), (9.3, 1.2), (9.8, 0).
c.
On a graph, points are at (0.4, 7.8), (3.1, 6.0), and (9.8, 0.5).
b.
On a graph, points are at (2, 7), (9.3, 1.2), (9.8, 1.5).
d.
On a graph, points are at (1.2, 7.2), (9.3, 1.2), (9.8, 0.2).
Please select the best answer from the choices provided
A
B
C
D
Answer: taake this link, it has all the answers
Step-by-step explanation: https://quizlet.com/183183758/statistical-studies-scatterplots-practiceamdm-flash-cards/
The best option for the points on the graph is points are at (0.4, 7.8), (3.1, 6.0), and (9.8, 0.5).
What are co ordinate axis?In two-dimensional Cartesian geometry, two intersecting straight lines are used as reference lines. In three-dimensional Cartesian geometry, three straight lines with a common point are the intersections of the three coordinate reference planes.
Estimation of the coordinates from the graph:From the attached file of scatter plot for the data, it is clear that the for every value of x there is a suitable value of y which was given in the question.
Considering x values and plot the y vale on the graph.
For x = 0.4; y = 7.8
For x = 3.1; y = 6.0
and for x = 9.8; y = 0.5
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Given any two squares, we can construct a square that equals (in area) the sum of the two given squares. Why?
We cannot construct a square that equals the sum of the areas of two given squares. This statement contradicts the mathematical principles and properties of squares and the Pythagorean theorem.
The statement that given any two squares, we can construct a square that equals the sum of the two given squares is actually false. This statement goes against the well-known mathematical concept known as the Pythagorean theorem.
The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem holds true for right-angled triangles, but it does not hold true for squares.
In fact, if we take two squares and try to add their areas together, the result will not be a square with an area equal to the sum of the two given squares. The resulting shape will be a non-square rectangle or some other irregular shape.
Therefore, we cannot construct a square that equals the sum of the areas of two given squares. This statement contradicts the mathematical principles and properties of squares and the Pythagorean theorem.
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A group of researchers is designing an experiment to test whether working late at night reduces a person's productivity. The researchers selected a sample of 60 adults, and over the next month, 30 of them will complete assigned tasks late at night whereas the remaining 30 will complete the same tasks during the afternoon. At the end of the month, the researchers will compare the average amount of time it takes each group to complete the tasks. Why is it important that the researchers use randomization to assign each of the adults to a group
Answer: Randomization prevents the researchers from specifically assigning all adults who regularly stay up late to complete assigned tasks late at night.
-Randomization ensures each group will be similar in everything except the time of day they are assigned to work.
-Randomization prevents the adults from selecting their own group.
Step-by-step explanation:
The options include:
a. Randomization ensures each group will be similar in everything except the time of day they are assigned to work.
b. Randomization prevents the adults from selecting their own group.
c. Randomization eliminates lurking variable from the experiment.
d. Randomization prevents the researchers from specifically assigning all adults who regularly stay up late to complete assigned tasks late at night.
e. Randomization ensures that there'll be an equal number of adults in each group.
Randomization refers to a method that is based on chance alone whereby the participants in a particular study are assigned to treatment group. Based on the information given, it us important that the researchers use randomization to assign each of the adults to a group because:
• Randomization ensures each group will be similar in everything except the time of day they are assigned to work.
• Randomization prevents the researchers from specifically assigning all adults who regularly stay up late to complete assigned tasks late at night.
• Randomization prevents the adults from selecting their own group.
Find the area to the following figure. Round to the one decimal place.
Answer:
91in^2
Step-by-step explanation:
First, add 11 and 15: 11 + 15 = 26
Second, divide it by 2: 26/2 = 13
Third, multiply it by the height: 13 * 7 = 91in^2
not sure what the answer is
Given:
Base dimensions of a rectangular prism = 15 cm × x cm
The height of the prism = 8 cm
Volume of the prism = 600 cm.
To find:
The value of x.
Solution:
Base area of the prism is:
[tex]B=length\times width[/tex]
[tex]B=15\times x[/tex]
[tex]B=15x[/tex]
Volume of the prism is:
[tex]V=Bh[/tex]
Where, B is the base area and h is the height of the prism.
Putting [tex]V=600,B=15x,h=8[/tex], we get
[tex]600=15x\times 8[/tex]
[tex]600=(8)(15)x[/tex]
[tex]\dfrac{600}{(8)(15)}=x[/tex]
Therefore, the correct option is C.
iGive a combinatorial proof of 1, 2+ 2.3+3.4++ (x - 1)^n = 2 =>("") Hint: Classify sets of three numbers from the integer interval [0...n] by their maximum element.
By classifying sets of three numbers from the interval [0...n] by their maximum element, we have provided a combinatorial proof of the identity 1, 2 + 2.3 + 3.4 + ... + (x - 1)^n = 2^(n+1).
The combinatorial proof of the identity 1, 2 + 2.3 + 3.4 + ... + (x - 1)^n = 2^(n+1) revolves around classifying sets of three numbers from the integer interval [0...n] by their maximum element.
Let's consider the right-hand side of the equation, which is 2^(n+1). This represents the number of subsets of an n-element set. We can think of each element in the set as having two choices: either it is included in a subset or not. Therefore, there are 2 choices for each element, resulting in a total of 2^(n+1) subsets.
Now, let's look at the left-hand side of the equation, which is the sum 1 + 2 + 2.3 + 3.4 + ... + (x - 1)^n. We can interpret each term as follows:
1: Represents the number of subsets with a maximum element of 0, which is only the empty set.
2: Represents the number of subsets with a maximum element of 1, which includes the subsets {0} and {1}.
2.3: Represents the number of subsets with a maximum element of 2, which includes the subsets {0, 1}, {0, 2}, and {1, 2}.
Similarly, for each subsequent term (x - 1)^n, it represents the number of subsets with a maximum element of x-1.
Now, if we add up all these terms, we are essentially counting the total number of subsets from the original set. This matches the right-hand side of the equation, which is 2^(n+1).
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use the linspace and plot commands in matlab to generate a figure containing the curves y=1.5sin(x) and y=x between x=0 and x=2.5
To generate a figure containing the curves y = 1.5*sin(x) and y = x in MATLAB using the linspace and plot commands, you can follow the steps below:
matlab
% Set the range of x values
x = linspace(0, 2.5, 100);
% Calculate y values for each curve
y1 = 1.5*sin(x);
y2 = x;
% Plot the curves
plot(x, y1, 'b', x, y2, 'r')
% Add labels and title
xlabel('x')
ylabel('y')
title('Curves: y = 1.5*sin(x) and y = x')
% Add a legend
legend('y = 1.5*sin(x)', 'y = x')
% Display the grid
grid on
In this code, we use linspace to create a range of x values from 0 to 2.5 with 100 points. Then, we calculate the corresponding y values for each curve using the equations y1 = 1.5*sin(x) and y2 = x. We use the plot command to plot the curves, with 'b' and 'r' specifying the colors of the curves. Next, we add labels, title, and a legend to the graph. Finally, we display the grid.
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The moment generating function for the standard normal distribution is given by My(t)=et. Use this MGF (and its derivatives) to show that the mean and variance of the standard normal distribution is 0
Using the MGF and its derivatives, we have shown that the mean and variance of the standard normal distribution are both 0.
The MGF for the standard normal distribution is given as:
M(t) = e^(t²/2)
To find the mean of the standard normal distribution, we take the first derivative of the MGF with respect to t and evaluate it at t = 0:
M'(t) = (1/2)e^(t²/2) × 2t
Evaluating at t = 0:
M'(0) = (1/2)e⁰ × 2(0) = 0
Since the first derivative of the MGF evaluated at t = 0 is 0, this implies that the mean of the standard normal distribution is 0.
To find the variance of the standard normal distribution, we take the second derivative of the MGF with respect to t and evaluate it at t = 0:
M''(t) = (1/2)e^(t²/2) × 2t² + (1/2)e^(t²/2)×2
Evaluating at t = 0:
M''(0) = (1/2)e⁰ × 2(0)² + (1/2)e⁰ × 2
= 0 + 1
= 1
Since the second derivative of the MGF evaluated at t = 0 is 1, this implies that the variance of the standard normal distribution is 1.
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What is the rate of change of the function described in
the table?
A. 12/5
B. 5
C. 25/2
D. 25
Answer:
B. 5
Step-by-step explanation:
This table does not describe a linear relationship. This is because the table does not have a constant rate of change; between the first two points, the function changes by
(1/2 - 1/10)/(0--1) = (5/10 - 1/10)/(1) = 5/10 - 1/10 = 4/10 = 2/5
Between the second two points, the function changes by
(5/2-1/2)/(1-0) = (4/2)/1 = 4/2 = 2
Comparing the first two y-coordinates, we have 1/10 and 1/2 = 5/10. This is the result of multiplication by 5.
Comparing the second two y-coordinates, we have 1/2 and 5/2. This is the result of multiplication by 5.
Comparing the rest of the y-coordinates, we can see that each time, the y-coordinate is multiplied by 5. This means the rate of change is 5.
Under her cell phone plan, Lily pays a flat cost of $40 per month and $4 per gigabyte.
She wants to keep her bill under $60 per month. Write and solve an inequality which
can be used to determine x, the number of gigabytes Lily can use while staying within
her budget.
Answer:
16
Step-by-step explanation:
60-44
The inequality is $40 + 4x < $60 and she has to spend 5 gigabytes to stay within her budget
Note that:
> means greater than
< means less than
≥ means greater than or equal to
≤ less than or equal to
The total amount Lily has to spend has to be less than $60
The total amount she can spend can be represented with this equation
Flat cost + variable cost < $60
Flat cost = $40
Variable cost = cost per gigabyte x gigabyte
$4 × x = $4x
$40 + 4x < $60
To solve for x, combine similar terms
$4x < $60 - $40
$4x < $20
x < 5
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Q1) Eruptions of the Old Faithful geyser in Yellowstone National Park typically last from 1.5 to 5 minutes. Between eruptions are dormant periods, which typically last from 50 to 100 minutes. A dormant period can also be thought of as the waiting time between eruptions. The duration in minutes for 40 consecutive dormant periods are given in the following table. 91 82 84 85 80 73 72 84 86 76 51 70 71 83 79 79 67 76 60 81 55 53 51 53 45 49 67 76 86 88 82 68 82 51 51 75 86 沙575 66 Assuming that the waiting time follows an Exponential distribution with mean parameter A, develop a uniformly most powerful test of size a = 0.01 for Hoλ2 80 vs H₁ A<80. Based on this test, draw a conclusion.
The duration in minutes for 40 consecutive dormant periods are given in the following table:91 82 84 85 80 73 72 84 86 76 51 70 71 83 79 79 67 76 60 81 55 53 51 53 45 49 67 76 86 88 82 68 82 51 51 75 86 575 66.
Assuming that the waiting time follows an exponential distribution with mean parameter A, a uniformly most powerful test of size α = 0.01 for H o λ^2=80 vs H1 A<80 can be developed as follows: The null and alternative hypotheses are as follows:H0:λ^2=80, that is, the mean of the exponential distribution is 80 squared.H1:A<80, which implies that the mean waiting time between eruptions is less than 80 squared.α=0.01 is the level of significance.
The following test statistic T is used: T = [n(λ^2-80)] / 80^2where n is the sample size, and the critical region is the left-tail rejection area. The probability of observing the values in the given sample or a more extreme set of values is calculated as follows: Since we are performing a one-tailed test, we divide α by 2.α/2 = 0.005
The area in the left tail is 0.005, and the corresponding z-score is -2.33.The null hypothesis is rejected if the computed value of the test statistic falls in the critical region, which is in the left-tail rejection region. T < -2.33
Since the test statistic T = -1.91 falls in the non-critical region, we fail to reject the null hypothesis at the α=0.01 level of significance. Therefore, based on this test, we can conclude that there is insufficient evidence to suggest that the mean waiting time between eruptions is less than 80 squared.
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Hey I'm Chloe Can you Help Me, I will give Brainlest, Thank you :)
During a professional baseball game, every spectator placed his or her ticket stub into one of several containers. After the game, the coach chose twenty people to march in the victory parade. What is the sample in this situation?
Answer:
The sample is the amount of people because not everyone is getting chosen
Step-by-step explanation:
I also agree, the coach choose certain people to march cuz not everyone is gonna be able to get used.
Select the proposition that is a tautology. a. (p ^ q) → p b.(p ∨ q) → p с. (р ^ q) → р d. (p ^ q) → p
The proposition that is a tautology is d. (p ^ q) → p. In a tautology, the truth value of the proposition is always true, regardless of the truth values of its individual components.
To determine if a proposition is a tautology, we can construct a truth table and evaluate all possible combinations of truth values for its variables.
For option d, (p ^ q) → p, we have the following truth table:
p q (p ^ q) (p ^ q) → p
T T T T
T F F T
F T F T
F F F T
The proposition that is a tautology is d. (p ^ q) → p.
In a tautology, the truth value of the proposition is always true, regardless of the truth values of its individual components. To determine if a proposition is a tautology, we can construct a truth table and evaluate all possible combinations of truth values for its variables.
For option d, (p ^ q) → p, we have the following truth table:
p q (p ^ q) (p ^ q) → p
T T T T
T F F T
F T F T
F F F T
As we can see, regardless of the truth values of p and q, the proposition (p ^ q) → p always evaluates to true. Therefore, option d is a tautology.
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What is the area of this polygon?
The middle is a square with side length of 12cm
Area of the square = 12 x 12 = 144 cm^2
There are 4 triangles with base of 13cm and height of 8 cm
Area of triangle = 1/2 x base x height
Area = 1/2 x 12 x 8 = 48cm^2 each
48 x 4 = 192 cm^2
Total area = 144 + 192 = 336 cm^2
Tommy is helping his mom at the grocery store. He notices that 5 pounds of potatoes cost $2.25. How much would 7 pounds of potatoes cost?
A. $3.45
B. $0.45
C. $3.15
D. $1.61
Answer:
Cost of 7 pound potato = $3.15
Step-by-step explanation:
Given:
Cost of 5 pound potato = $2.25
Find:
Cost of 7 pound potato
Computation:
Cost of 1 pound potato = 2.25 / 5
Cost of 1 pound potato = $0.45
Cost of 7 pound potato = 7 x Cost of 1 pound potato
Cost of 7 pound potato = 7 x 0.45
Cost of 7 pound potato = $3.15
Does any one know the answer to this thank you
Answer:
The First choice
Step-by-step explanation:
Given the differential equation
dy/dx = 3xy+y^2 / x^2+xy
i) Show that this equation is homogeneous.
ii) By substituting y =xv, solve this differential equation with initial
condition y(1) = 4.
The given differential equation is shown to be homogeneous, and the solution to the equation with the initial condition y(1) = 4 is :
y = 6x^2 * e^(-x/2) - 2x.
i) To show that the differential equation is homogeneous, we need to verify that it is invariant under the transformation y = ux, where u is a function of x.
Let's substitute y = ux into the given differential equation:
dy/dx = 3xy + y^2 / x^2 + xy
Using the chain rule, we can express dy/dx in terms of u and x:
dy/dx = d(ux)/dx = u + x * du/dx
Substituting this into the differential equation:
u + x * du/dx = 3x(ux) + (ux)^2 / x^2 + x(ux)
Simplifying the equation:
u + x * du/dx = 3u + u^2 / x + u
The equation can be further simplified:
x * du/dx = 2u + u^2 / x
We can see that the resulting equation is independent of x. Hence, the original differential equation is homogeneous.
ii) To solve the homogeneous differential equation, let's substitute y = xv back into the equation:
x * du/dx = 2u + u^2 / x
Multiplying through by x:
x^2 * du/dx = 2xu + u^2
Rearranging the equation:
x^2 * du / (2u + u^2) = dx
We can now integrate both sides:
∫ x^2 * du / (2u + u^2) = ∫ dx
The left-hand side can be further simplified using partial fraction decomposition:
∫ (A/u + B/(u+2)) du = ∫ dx
Solving for A and B, we get:
A = -2, B = 1
Substituting back into the integral:
∫ (-2/u + 1/(u+2)) du = ∫ dx
Simplifying the integral:
-2ln|u| + ln|u+2| = x + C
Now substituting u = y/x:
-2ln|y/x| + ln|(y/x)+2| = x + C
Using properties of logarithms, we can simplify this equation further:
ln((y+2x)/x^2) = -x/2 + C
Taking the exponential of both sides:
(y+2x)/x^2 = e^(-x/2+C)
Simplifying the right-hand side by combining e^C into a constant A:
(y+2x)/x^2 = A * e^(-x/2)
Now, solving for y:
y + 2x = Ax^2 * e^(-x/2)
Finally, rearranging the equation to solve for y:
y = Ax^2 * e^(-x/2) - 2x
Given the initial condition y(1) = 4, we can substitute x = 1 and y = 4 into the equation:
4 = A * e^(-1/2) - 2
Solving for A:
A * e^(-1/2) = 6
A = 6 * e^(1/2)
Substituting the value of A back into the equation, we have:
y = 6x^2 * e^(-x/2) - 2x
So the solution to the differential equation with the initial condition y(1) = 4 is y = 6x^2 * e^(-x/2) - 2x.
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Find the general solution of the given differential equation. 7 dy dx + 56y = 8
y(x) =
Give the largest interval I over which the general solution is defined. (Think about the implications of any singular points. Enter your answer using interval notation.)
Determine whether there are any transient terms in the general solution.
The general solution of the given differential equation 7dy/dx + 56y = 8 is y(x) = -x/8 + C e^(-8x/7), where C is a constant.
To solve the differential equation, we first rearrange it to isolate dy/dx: dy/dx = (8 - 56y)/7. This is a first-order linear differential equation. The integrating factor is e^(∫(-56/7)dx) = e^(-8x/7). Multiplying both sides of the equation by this integrating factor, we obtain e^(-8x/7) dy/dx + 8e^(-8x/7)y = 8e^(-8x/7). The left-hand side can be written as the derivative of y multiplied by e^(-8x/7). Integrating both sides gives ∫d(y e^(-8x/7)) = ∫8e^(-8x/7) dx. Solving these integrals and rearranging, we find the general solution y(x) = -x/8 + C e^(-8x/7), where C is the constant of integration.
The largest interval I over which the general solution is defined is (-∞, ∞) since there are no singular points or restrictions mentioned in the differential equation. This means that the solution is valid for all real values of x.
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In which interval is the radical function f of x is equal to the square root of the quantity x squared plus 2 times x minus 15 end quantity increasing?
[3, [infinity])
(4, [infinity])
[–5, 3]
(–[infinity], –5] ∪ [3, [infinity])
The correct answer is, [–5, 3]. In the other words, the interval in which the function [tex]f(x) = \sqrt{x^2 + 2x - 15}[/tex] is increasing is [–5, 3].
To determine the interval in which the radical function [tex]f(x) = \sqrt{x^2 + 2x - 15}[/tex] is increasing, we need to find the interval(s) where the derivative of the function is positive.
Let's first find the derivative of f(x):
[tex]f'(x) = (1/2) * (x^2 + 2x - 15)^(-1/2) * (2x + 2)[/tex]
To find where f'(x) > 0, we set f'(x) = 0 and solve for x:
[tex](1/2) * (x^2 + 2x - 15)^(-1/2) * (2x + 2) = 0[/tex]
Since the derivative is never equal to zero (since the denominator (x^2 + 2x - 15)^(-1/2) is never equal to zero), there are no critical points.
To determine the intervals of increase, we can evaluate f'(x) at test points in each interval. We'll consider the intervals defined by the given answer choices:
[3, ∞):
Choose a test point x > 3, let's say x = 4.
Evaluate [tex]f'(4) = (1/2) * (4^2 + 24 - 15)^{(-1/2)} * (24 + 2)[/tex]
[tex]= (1/2) * (16 + 8 - 15)^{(-1/2)} * 10[/tex]
[tex]= (1/2) * (9)^{(-1/2)} * 10[/tex]
= (1/2) * (1/3) * 10
= 5/3
Since f'(4) > 0, the function is increasing in the interval [3, ∞).
(4, ∞):
Choose a test point x > 4, let's say x = 5.
Evaluate f'(5) = (1/2) * (5^2 + 25 - 15)^(-1/2) * (25 + 2)
= (1/2) * (25 + 10 - 15)^(-1/2) * 12
= (1/2) * (20)^(-1/2) * 12
Since f'(5) = 0, the function is not increasing in the interval (4, ∞).
[–5, 3]:
Choose a test point x in the interval, let's say x = 0.
Evaluate [tex]f'(0) = (1/2) * (0^2 + 20 - 15)^{(-1/2)} * (20 + 2)[/tex]
[tex]= (1/2) * (-15)^{-1/2} * 2[/tex]
[tex]= (1/2) * (1/\sqrt{15}) * 2[/tex]
[tex]= 1/\sqrt{15}[/tex]
Since f'(0) > 0, the function is increasing in the interval [–5, 3].
(–∞, –5] ∪ [3, ∞):
Since we have already determined the function is increasing in [–5, 3] and [3, ∞), this interval is valid.
Therefore, the correct answer is, [–5, 3]. In the other words, the interval in which the function [tex]f(x) = \sqrt{x^2 + 2x - 15}[/tex] is increasing is [–5, 3].
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If you move a decimal to the left 3 times will the numbers increase in value
Answer:
The number will decrease in value
Step-by-step explanation: For example if we had 638.23 if I move the decimal to the left 3 times it would be .63823, and if we add zeros it would look like this, 0.63823 so the number would decrease in value
Lindsay and Lorraine are trying to match the jump rope world record. Together, they need to jump 48 times in a row. Lindsay has gotten 14 jumps in a row, and Lorraine has gotten 13. Write an equation using (j) as the variable and show your work to determine how many more jumps they need to complete.
( if u steal my points ill steal yours)
(no links or ill report)
Answer: They need to complete 21 more jumps all together .
Step-by-step explanation: So 14+13=J
14+13=27
27=j
World record: 48; 48-27=21
They will have to jump 21 more times in a row .
What is the solution to the equation?
O
a+5=9
a+53 -9
a+52 +53-9-5
a=3
a+
a+53-9
O a+52 +53 - 9+53
-
a+53-9
3
|
a = 14 3
6
2
Answer:
The format of this equation is wrong
It takes me into 30 minutes to walk from home to school when walking at 5 km per hour What is her average cycling speed If it takes her 15 minutes by bike to travel the same distance
Answer:
10 km/h
Step-by-step explanation:
I'm not really sure about this but no ones answering your question and I wanna help.
So basically to calculate the average speed you need to divide the distance travelled by time taken
But you do not have the distance traveled. But it is mentioned that it takes u 30 minutes to walk from home to school when walking at 5 km/h so to find the distance all you have to do is... 30 x 5 = 150 km
Now that we have the time and distance all we have to do is find the average speed.
Average Speed = distance ÷ time
So 150 ÷ 15 = 10 km/h
HELP ASAP SKAKSKKAMAAAAA
Answer:
(240,20)
Step-by-step explanation:
Find the center of the ellipse.
x2 + 4y2 – 10x – 40y + 121 = 0
Answer:
i dont what an ellipse is but here's the answer:
8x + 32y = 121
Answer:
123‐10×40y=0
10×+40y=123
Shoulda let me have you i coulda made you so happy but ion do 2nd chances, forever i wish u happiness PERIODT DOE.
now its time to make bankk
Answer:
periodttt. get out ya bag n make det money up.
Step-by-step explanation:
brainliestt:)?
whats 38 divided by 70
Answer:
.5428571 repeating
or 19/35
PLEASE HELP I HAVE TO TURN THIS IN IN 1 HOUR!
A round clock on a classroom wall has a diameter of 12 inches. What is the approximate area of the clock?
A. 38 square inches
B. 113 square inches
C. 226 square inches
D. 452 square inches
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