Answer:
It is 7
Step-by-step explanation:
Describe a quadrillateral and a hexagon by the number of vertices it has.
The main difference between Hexagon and Quadrilateral is that the Hexagon is a polygon with six sides and Quadrilateral is a polygon with four sides. In geometry, a hexagon (from Greek ἕξ hex, "six" and γωνία, gonía, "corner, angle") is a six sided polygon or 6-gon.
Can someone help me pls I’m stuck on both problems
Answer:
171
Step-by-step explanation:
6x6x4.75 is 171
please help me do it
-3(w+1)+8(w-3)=-5w+9+6w
PLEASE HURRY AND ANSWER
Estimate the area under the graph of f(x) =10 sqrt x from x = 0 to x = 4 using four approximating rectangles and right endpoints. (Round your answers to four decimal places.)
(a) Use four approximating rectangles and right endpoints.
R4=______________________
(b) Use four approximating rectangles and left endpoints.
L4=_______________________
(a) R4= 61.4626 and (b) L4= 61.4626.
In order to estimate the area under the graph of f(x) = 10 sqrt x from x = 0 to x = 4 using four approximating rectangles and right endpoints, we need to use the formula:
Rn = Δx [f(x1) + f(x2) + f(x3) + ... + f(xn)], where Δx is the width of each rectangle and f(xi) is the height of the rectangle at the right endpoint of the ith interval.
Step 1: Calculation of ∆x.∆x = (4 - 0)/4 = 1
Step 2: Calculation of xi for i = 1, 2, 3 and 4.x1 = 1, x2 = 2, x3 = 3, x4 = 4
Step 3: Calculation of f(xi) for i = 1, 2, 3 and 4.f(x1) = 10√(1) = 10f(x2) = 10√(2) ≈ 14.1421f(x3) = 10√(3) ≈ 17.3205f(x4) = 10√(4) = 20
Step 4: Calculation of R4.R4= ∆x [f(x1) + f(x2) + f(x3) + f(x4)]R4= 1[10 + 14.1421 + 17.3205 + 20]= 61.4626Area ≈ 61.4626 square units.
Step 5: Calculation of L4.Ln = ∆x [f(x0) + f(x1) + f(x2) + ... + f(xn-1)]
Where x0 is the initial value.
Here, we can find the value of L4 by using the left endpoints.
So, x0 = 0L4 = ∆x [f(x0) + f(x1) + f(x2) + f(x3)]L4 = 1 [f(0) + f(1) + f(2) + f(3)]L4 = 1 [10 + 14.1421 + 17.3205 + 20]L4 = 61.4626
Therefore, (a) R4= 61.4626 and (b) L4= 61.4626.
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QR Factorization is a useful technique when the normal equations
for a least squares problem are
ill-conditioned. What does ill-conditioned
mean?
(Limit your answer to 25 words of less.)
Ill-conditioned means that the problem or system being considered is sensitive to small changes in the input or data, leading to unstable or inaccurate results.
When solving a least squares problem using the normal equations, ill-conditioning refers to situations where the matrix involved is nearly singular or has a high condition number.
This means that small perturbations or errors in the data can result in large changes in the computed solution.
In the context of QR factorization, if the normal equations for a least squares problem are ill-conditioned, it implies that the matrix being decomposed using QR factorization is close to being singular or has a high condition number. QR factorization can help in such cases by providing a more stable and accurate solution compared to directly solving the normal equations.
QR factorization decomposes a matrix into the product of an orthogonal matrix Q and an upper triangular matrix R. This factorization can help mitigate the effects of ill-conditioning by providing a numerically stable way to solve the least squares problem.
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(a) For the same data and null hypothesis, is the P-value of a one-tailed test (right or left) larger or smaller than that of a two-tailed test? Explain.
The P-value for a one-tailed test is larger because the two-tailed test includes the area in both tails. The P-value for a one-tailed test is smaller because the two-tailed test includes the area in only one tail. The P-value for a one-tailed test is smaller because the two-tailed test includes the area in both tails. The P-value for a one-tailed test is larger because the two-tailed test includes the area in only one tail.
The P-value of a one-tailed test is generally smaller than that of a two-tailed test when testing the same null hypothesis and using the same level of significance.
The P-value of a one-tailed test is generally smaller than that of a two-tailed test when testing the same null hypothesis and using the same level of significance. This is because a one-tailed test focuses on a specific direction of the hypothesis, while a two-tailed test considers both directions.
In a one-tailed test, the null hypothesis is rejected only if the test statistic falls in the critical region in one direction. For example, if the null hypothesis is that a mean is less than or equal to a certain value, the critical region will be in the lower tail of the distribution. Therefore, the probability of obtaining a test statistic in the critical region is smaller compared to a two-tailed test, where the critical region is split between both tails of the distribution.
As a result, the P-value of a one-tailed test is smaller than that of a two-tailed test, given the same null hypothesis and level of significance. However, it's important to note that the choice between a one-tailed or two-tailed test should be based on the specific research question, rather than the desire for a smaller P-value.
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can someone help me I’m stuck on this and can’t figure it out
Answer:
20%
Step-by-step explanation:
Answer:
20%
Step-by-step explanation:
19.99 - 3.998(20%) = 15.992 which when rounded is 15.99
can someone please help me pls
Answer:
The answer is 21 units²
Step-by-step explanation:
To find the area of a right triangle you use the formula A=ab/2
So
A=6×7/2 = 21
Find a primitive root, for all positive integral m, modulo each integer below. (a) 7 (Hint: Using Corollary 5.15, find a common primitive root r modulo 7 and 72. The proof of Proposition 5.17 then guarantees that r is a primitive root modulo 7" for all positive integral m.) (b) 11m (c) 13m (d) 17m
To find a primitive root modulo a given integer, we can use Corollary 5.15 and Proposition 5.17. For (a) 7, a common primitive root r can be found by using these results. Similarly, for (b) 11m, (c) 13m, and (d) 17m, the same approach can be applied to find primitive roots modulo each integer.
(a) To find a primitive root modulo 7 and 72, we can use Corollary 5.15, which states that if r is a primitive root modulo n, then r is also a primitive root modulo any power of n. Since 7 is a prime number, we can easily find a primitive root modulo 7. Let's say r is a primitive root modulo 7. Using Proposition 5.17, which guarantees that a primitive root modulo a prime number remains a primitive root modulo any positive integral power of the prime, we can conclude that r is a primitive root modulo 7 for all positive integral m.
(b) Similarly, for 11m, we can find a primitive root modulo 11 and use Proposition 5.17 to prove that it is a primitive root modulo 11m for all positive integral m.
(c) The same approach can be applied to find a primitive root modulo 13m.
(d) Finally, for 17m, we can find a primitive root modulo 17 and use Proposition 5.17 to prove its primitiveness modulo 17m for all positive integral m.
By using Corollary 5.15 and Proposition 5.17, we can find a common primitive root for all the given integers modulo their corresponding powers.
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A parallelogram has four sides that are the same length.is a it square ?
Answer:
I'm pretty sure no because a square always has 4 right angles and all sides are the same length
the parallelogram could possibly be a rectangle tho
I could be completely wrong but I hope this helps
Help me please!!!!!!
(1-x)²-(-x-2)×(2x+3)
can someone explain how to solve this step by step please
algebra if x/y = 5/8 which of the following must be true?
Consider this situation: A school publicizes that the proportion of attending students who are involved in at least one extracurricular activity is 70% Would we employ a two-tailed test or a one-tailed test to test the claim about the proportion of students involved in extracurricular activities? Chi-squared (one tailed) two-tailed test O Chi-squared (two tailed) One-tailed
After considering the given data we conclude that one tailed test can be utilised for this claim.
To test the claim about the proportion of students involved in extracurricular activities, we would employ a one-tailed test.
A one-tailed test is applied when the alternative hypothesis is directional, which projects the it predicts the direction of the difference between the sample proportion and the claimed proportion. For the given case, the alternative hypothesis will be that the proportion of attending students involved in at least one extracurricular activity is greater than 70%.
Hence, we would use a one-tailed test to test this claim.
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A researcher checks different times that students take to complete an IQ test. The Samples are from different states. Please check if their means are significantly different or the difference is just by chance. Significance level is 5%. The premise is that the ANOVA assumption of normality and homogeneity of variance has been satisfied.
If we fail to reject the null hypothesis, we cannot conclude that the means are significantly different.
To check whether the means of the samples from different states are significantly different or if the difference is just by chance, we can use the one-way analysis of variance (ANOVA).
The premise is that the ANOVA assumption of normality and homogeneity of variance has been satisfied.
Here, the null hypothesis is that the means of all the samples are equal while the alternative hypothesis is that at least one mean is different.
The significance level is 5%.
To perform the ANOVA test, we can follow these steps:
Step 1: State the null and alternative hypotheses
H0: μ1 = μ2 = μ3 = ... = μk (where k is the number of groups)
Ha: At least one mean is different
Step 2: Set the significance level (α)α = 0.05
Step 3: Calculate the test statistic
The test statistic used in ANOVA is the F-statistic, which is the ratio of the between-group variance to the within-group variance.
If the F-statistic is large and the p-value is less than the significance level, we can reject the null hypothesis and conclude that at least one mean is different. If the F-statistic is small and the p-value is greater than the significance level, we fail to reject the null hypothesis and conclude that the means are not significantly different.
Step 4: Calculate the p-value
Using the F-statistic and the degrees of freedom for the between-group and within-group variance, we can calculate the p-value. If the p-value is less than the significance level, we can reject the null hypothesis. If the p-value is greater than the significance level, we fail to reject the null hypothesis.
Step 5: State the conclusion
Based on the p-value, we can either reject or fail to reject the null hypothesis. If we reject the null hypothesis, we can conclude that at least one mean is different. If we fail to reject the null hypothesis, we cannot conclude that the means are significantly different.
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Find the surface area.
24 in.
40 in.
10 in.
26 in.
Answer:
24 IN.
Sorry for all caps
Step-by-step explanation:
PLEASE PLEASE PLEASE HELP
WITH 2D AND 3A ONLY, PLEASE SHOW YOUR WORKING
Answer:
2d) d = 60°
3a) x = 50°
Step-by-step explanation:
2d) both of the diagonals pass through the centre of the circle and divide the rectangle into 4 triangles,then it means that;
30 + d = 90°
d = 90 - 30
d = 60°
3a) since O is the centre of the circle, and we can see that the 2 chords directly above and below the centre are parallel to each other, it means that angle x = 50° since the two sides projecting from the centre to form x° and 50° are equal and in essence used to form isosceles triangle.
331 student went on a field trip. Six bises were filled and 7 students traveled in cars. How many students were in each bus?
Answer:
54 students
Step-by-step explanation:
if 7 of 331 students did not go on buses then 331 - 7 = 324 went on buses.
324 divided by 6 = 54 students
on the 3-days of their vacation the Davis family traveled 417 mi 45.3 mi and 366.9 mi how far did they travel all together
Answer:
829.2 miles
Step-by-step explanation:
To find how far they traveled in total, add together all of the distances:
417 + 45.3 + 366.9
= 829.2
So, all together, they traveled 829.2 miles
Isabel and Jonah had 2 pies. Isabel wrote the equation ½ + ⅙ = 4/6 and Jonah wrote 3/6 + 1/6 = 4/6 to represent combining the pie pieces. Explain why both equations are correct.
PLEASE JUST TYPE THE ANSWER AND DON’T LEAD TO ANY LINKS
Slove the system of linear equations by graphing y=-x+7 y=x-1
Answer:
no
Step-by-step explanation:
what percent of 15 is 0.6?
Answer:
4 is a percent of 15 is 0.6
Step-by-step explanation:
if you out 15/0.6 as a fraction you will get 4
Answer:
4%
Step-by-step explanation:
STEP 1: Put the numbers into a fraction
0.6/15
STEP 2: Divide
0.6/15 = 0.04
STEP 3: Move decimal point two places to the right
0.04 -> 4
Answer: 4%
UNIT 6: CEREAL BOX PROJECT / PORTFOLIO
A company has released the following possible designs for a new cereal box. The cost of production depends upon the cost of materials, which is determined by how much material is used. The amount of material used is equal to the surface area of the package plus some overlap. The amount of product contained in the package is the volume of the packaging.
Find the surface area and volume of the following possible cereal boxes. Determine the cost of making the cereal box assuming that cardboard costs $0.05 per square inch.
Answer:
DO you go to a connexus school? I do and i have this same project im trying to find answers to!
Step-by-step explanation:
For the first box the volume, surface area, and cost of materials are 180 cubic inches, 258 square inches, and $12.9, for the second box 192 cubic inches, 224.4 square inches, and $11.22 for the third box 235.61 cubic inches, 227.76 square inches, and $11.38.
What is volume?It is defined as a three-dimensional space enclosed by an object or thing.
For the first cereal box:
Volume = L×W×H
Here L = 7.5 in, W = 2 in, and H = 12 in
Volume = 7.5×2×12 = 180 cubic inches
Surface area = 2(L×B+B×H+H×L)
= 2(7.5×2+2×12+12×7.5)
= 2(15+24+90)
= 258 square inches
Cost for this cereal box = 258×0.05 = $12.9
For the second cereal box:
[tex]\rm Volume = \frac{1}{3} bh[/tex]
b is the area and h, is the height of the pyramid.
b = 8×6 = 48 square inches, h = 12 inches
[tex]\rm Volume = \frac{1}{3} \times48\times12[/tex]
Volume = 192 cubic inches
[tex]\rm Surface \ area = b+\frac{1}{2} ps[/tex]
p is the perimeter of the base = 2(8+6) = 28 in
Slant height s = 12.6 in
[tex]\rm Surface \ area = 48+\frac{1}{2} (28)(12.6)[/tex] = 224.4 square inches
Cost of this cereal box = 224.2×0.05 = $11.22
For the third box:
Volume = πr²h
r = 2.5 in and h = 12 in
Volume = π(2.5)²(12) = 235.61 cubic inches
Surface area = 2πr(h+r) = 2π(2.5)(12+2.5) = 227.76 square inches
Cost of this cereal box = 227.76×0.05 = $11.38
Thus, for the first box the volume, surface area, and cost of materials are 180 cubic inches, 258 square inches, and $12.9, for the second box 192 cubic inches, 224.4 square inches, and $11.22 for the third box 235.61 cubic inches, 227.76 square inches, and $11.38.
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Dwayne buys ingredients to make a cake. He buys 1 1/2 pounds of flour, 12 ounces of coconut, and 1 1/4 pounds of sugar. What is the total weight of the ingredients Dwayne bought?
Answer:
56 oz or 3 1/2 lbs
Step-by-step explanation:
1 1/4 + 1 1/2 = 2 3/4lbs = 44 oz
44 oz + 12 oz = 56 oz
Answer:
81/4
Step-by-step explanation:
the answer is not sure
Discrete math
Euclidean Algorithm (a) Find the Greatest Common Divisor of 27,720 and 58,212 (b) Find integers r and s so that 27720-5 + 58212.5 = GCD(27720, 58212) =
a) The GCD of 27,720 and 58,212 is 1.
b) Integers r = 3 and s = -1 satisfy the equation 27720r + 58212s = GCD(27720, 58212).
How to find the greatest common divisor (GCD) of 27,720 and 58,212?(a) To find the greatest common divisor (GCD) of 27,720 and 58,212, we can use the Euclidean Algorithm.
Divide 58,212 by 27,720.
58,212 ÷ 27,720 = 2 remainder 2
Divide the previous divisor (27,720) by the remainder (2).
27,720 ÷ 2 = 13,860
Divide the previous remainder (2) by the new remainder (2).
2 ÷ 2 = 1
Since the remainder is now 1, the GCD is found.
Therefore, the greatest common divisor (GCD) of 27,720 and 58,212 is 1.
How to find integers r and s such that 27720r + 58212s = GCD(27720, 58212)?(b) To find integers r and s such that 27720r + 58212s = GCD(27720, 58212), we can use the Extended Euclidean Algorithm.
Using the Euclidean Algorithm from part (a), we can backtrack to find the coefficients r and s:
2 = 27,720 - 13,860
Substituting the values:
2 = 27,720 - (58,212 - 2 * 27,720)
Simplifying:
2 = 3 * 27,720 - 58,212
Comparing with the form 27720r + 58212s = GCD(27720, 58212):
r = 3
s = -1
Therefore, we substitute r = 3 and s = -1 into the equation 27720r + 58212s, it will result in the greatest common divisor (GCD) of 27720 and 58212.
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Recall: Sampling Distributions Consider any population. Then for any n ,the sampling distribution of the sample mean will have mean Mg = Hy and standard deviation of a Consider a population that is N(Hz, Ox). Then for any n, the sampling distribution of the sample mean is normally distributed with mean Hz = Hly and standard deviation of x • . Central Limit Theorem (CLT): Consider any population with mean My and standard deviation Oy. Then for n large (n 2 30), the sampling distribution of the sample mean is approximately normal with mean Hz = Hly and standard deviation ох x √n = 3. 1) A company making electronic equipment experiences a production stoppage on average of one time per month. Assume the number of stoppages per month can be modeled according to a random variable X- ~ POIS (1) a) Complete the following table for this random variable. PARAMETERS Notation Numerical Value Mean Variance Standard Deviation
The Poisson distribution with a parameter λ = 1 accurately models the production stoppages, where on average, the company experiences one stoppage per month with a relatively small amount of variability.
The scenario describes a company's production stoppages, which can be modeled using a Poisson distribution with a parameter (mean) of λ = 1. In a Poisson distribution, the mean, variance, and standard deviation are all equal.
The mean (μ) represents the average number of stoppages per month, which in this case is 1. This means that, on average, the company experiences one production stoppage per month.
The variance (σ^2) also has a value of 1 in a Poisson distribution. It measures the spread or variability of the data around the mean. In this case, the variance of 1 indicates that there is some fluctuation in the number of stoppages, but it is relatively small.
The standard deviation (σ) is equal to the square root of the variance, which is also 1 in this scenario. It represents the average amount of deviation from the mean. A standard deviation of 1 suggests that most of the observations will be within one unit of the mean.
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an icecube has a mass of 18g and a volume of 20cm3 what is the density
Answer:
.9 g/cm^3
Step-by-step explanation:
density formula is mass divided by volume.
d=m/v
d=18g/20cm^3
d=0.9
Consider the function f(x) = 25 - x ^ 2
(a) Use a Riemann sum to estimate the area under the graph of between x = - 3 and x = 5 Divide the interval [-3, 5] into 4 subintervals each of the same length by using left-hand and midpoint approximation. Sketch the 4 rectangles that approximates the area under the curve.
(b) Use the limit of a Riemann sum to find the exact area of the region between the curve
y= f(x) and the x-axis on the interval [-3,5].
(a) The area under the curve is 154 square units.
(b) The exact area of the region between the curve y= f(x) and the x-axis is (118 / 3) square units.
(a) The given function is f(x) = 25 - x² .
We need to estimate the area under the graph between x = - 3 and x = 5 by dividing the interval [-3, 5] into 4 subintervals each of the same length by using left-hand and midpoint approximation and sketch the 4 rectangles that approximates the area under the curve.
The width of each rectangle is given by Δx, where Δx = (b - a) / n = (5 - (-3)) / 4 = 2.
The height of each rectangle is determined by either left-hand approximation or midpoint approximation.
1. Left-hand approximation: In the left-hand approximation method, the height of each rectangle is taken from the left endpoint of each subinterval. We have:
Left endpoint of the 1st subinterval is x₁ = -3 Left endpoint of the 2nd subinterval is x₂ = -1 Left endpoint of the 3rd subinterval is x₃ = 1 Left endpoint of the 4th subinterval is x₄ = 3
Thus, the heights of the four rectangles are: f(x₁) = f(-3) = 16f(x₂) = f(-1) = 24f(x₃) = f(1) = 24f(x₄) = f(3) = 16
We sketch the four rectangles as follows:
The total area of the four rectangles is the sum of the individual areas of the rectangles.
We have: Area ≈ [f(-3) + f(-1) + f(1) + f(3)] Δx= [16 + 24 + 24 + 16] × 2= 80 square units.2.
Midpoint approximation: In the midpoint approximation method, the height of each rectangle is taken from the midpoint of each subinterval.
We have: Midpoint of the 1st subinterval is x₁* = -2 Midpoint of the 2nd subinterval is x₂* = 0 Midpoint of the 3rd subinterval is x₃* = 2 Midpoint of the 4th subinterval is x₄* = 4
Thus, the heights of the four rectangles are: f(x₁*) = f(-2) = 21f(x₂*) = f(0) = 25f(x₃*) = f(2) = 21f(x₄*) = f(4) = 9
We sketch the four rectangles as follows:
The total area of the four rectangles is the sum of the individual areas of the rectangles.
We have:
Area ≈ [f(-2) + f(0) + f(2) + f(4)] Δx= [21 + 25 + 21 + 9] × 2= 154 square units.
(b) The exact area of the region between the curve y = f(x) and the x-axis on the interval [-3, 5] is given by the limit of a Riemann sum as the number of subintervals n approaches infinity.
We have:
Area = ∫[(-3, 5)] f(x) dx= ∫[-3, 5] (25 - x²) dx
= [25x - (x³ / 3)]|[-3, 5]
= [125 - (125 / 3)] - [-75 + (27 / 3)]
= (100 / 3) + (18 / 3)
= (118 / 3) square units.
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Method begins with a matrix or table describing the characteristics of a target population? a. probability b. snowball c. quota d. simple random sampling
The method that begins with a matrix or table describing the characteristics of a target population is quota sampling.
Quota sampling is a non-probability sampling method in which the researcher first creates a matrix or table describing the characteristics of the target population. Once the matrix is created, the researcher then selects a certain number of people from each category in the matrix until the desired sample size is reached.
For example, if the researcher is interested in studying the political views of college students, they might create a matrix with the following categories:
Gender: Male, Female
Age: 18-21, 22-25, 26-30
Political affiliation: Democrat, Republican, Independent
Once the matrix is created, the researcher might then select 100 students from each category, for a total sample size of 300.
Quota sampling is a quick and easy way to obtain a sample of a population. However, it is not a probability sampling method, which means that the sample may not be representative of the population. This is because the researcher is not randomly selecting participants from the population. Instead, they are selecting participants based on their characteristics.
As a result, quota sampling can lead to bias in the results of a study. For example, if the researcher selects more students from one political affiliation than another, the results of the study may be biased in favor of that political affiliation.
Overall, quota sampling is a useful tool for researchers who need to obtain a sample of a population quickly and easily. However, it is important to be aware of the limitations of this method and to take steps to minimize bias in the results of the study.
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