HELP!!!! Michelle is having a poster printed from a picture that is 4 inches wide by 6 inches tall. The enlarged poster is 20 inches wide and 30 inches tall. Which proportion can be used to represent the picture sizes?
StartFraction 6 over 4 EndFraction = StartFraction 20 over 30 EndFraction
StartFraction 4 over 6 EndFraction = StartFraction 20 over 30 EndFraction
StartFraction 4 over 6 EndFraction = StartFraction 30 over 20 EndFraction
StartFraction 4 over 30 EndFraction = StartFraction 6 over 20 EndFraction

The linear combination of f1, f2, and f3 can be expressed as: v = −1 f1 + 1 f2 − 2 f3.

Since f1, f2, and f3 form an orthogonal basis of ℝ3, we can express any vector in ℝ3 as a linear combination of these three vectors. Let's find the coefficients of the linear combination for v.

Let a, b, and c be the coefficients of f1, f2, and f3, respectively, in the linear combination. Then we have:

v = a f1 + b f2 + c f3

Substituting the given values, we get:

0 −9 0 = a (−1 −2 −2) + b (−2 −1 2) + c (−2 2 −1)

Simplifying this equation, we get a system of three linear equations:

−a − 2b − 2c = 0

−2a − b + 2c = −9

−2a + 2b − c = 0

Solving this system of equations, we get:

a = −1, b = 1, c = −2

Therefore, we can express v as a linear combination of f1, f2, and f3 as:

v = −1 f1 + 1 f2 − 2 f3

In other words, v is orthogonal to f1, f2, and f3, and can be expressed as a linear combination of them.

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Answer:

x=3 y=1

Step-by-step explanation:

plug in y. 2x-3(7-2x)=3

then get 8x=24

x=3

then plug in x in either equation

y=7-2(3)

y=1

The 80% confidence interval for the given pair of observations is 3. The 98% confidence interval for the given pair of observations is (1.02, 6.98).

The formula to calculate the 80% confidence interval for the given pair of observations is given as follows:Lower limit = Y - Zc/2(σ/√n)Upper limit = Y + Zc/2(σ/√n)where Y is the mean value of all the observations, σ is the standard deviation of all the observations, n is the sample size, and Zc is the critical value of Z at 10% significance level.From the given pair of observations, the mean is 4. The standard deviation is 1.414, which is calculated as the square root of the variance of all the observations (Variance = Σ (Xi - Mean)² / n)Thus, using the formula, we can calculate the 80% confidence interval as follows:Lower limit = 4 - (1.2816 * 1.414 / √3) = 2.18Upper limit = 4 + (1.2816 * 1.414 / √3) = 5.82The 80% confidence interval for the given pair of observations is (2.18, 5.82)

The formula to calculate the 98% confidence interval for the given pair of observations is given as follows:Lower limit = Y - Zc/2(σ/√n)Upper limit = Y + Zc/2(σ/√n)where Y is the mean value of all the observations, σ is the standard deviation of all the observations, n is the sample size, and Zc is the critical value of Z at 1% significance level.From the given pair of observations, the mean is 4. The standard deviation is 1.414, which is calculated as the square root of the variance of all the observations (Variance = Σ (Xi - Mean)² / n)Thus, using the formula, we can calculate the 98% confidence interval as follows:Lower limit = 4 - (2.3263 * 1.414 / √3) = 1.02Upper limit = 4 + (2.3263 * 1.414 / √3) = 6.98The 98% confidence interval for the given pair of observations is (1.02, 6.98).

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Answer:

24t^2 + 11

Step-by-step explanation:

There is a lot left out of this question. If you are allowed rational numbers, a common factor could be 24.

24(t^2 + 11/24) is one possibility.

I think what you mean is there an integer that is a common factor. There is not.

So the answer is 24t^2 + 11

There is one more thing that needs to be clarified. Can you give t a value? If you can, then t = 11

the common factor would be

11(24 * 11 + 1)

I doubt very much that this is what is meant, but the question really should make it clear.

Answer:

1(24f²+11) is your answer

Step-by-step explanation:

24f2+11

taking common

1(24f²+11)

Pythagoras theorem states that in a right angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. We can represent the length of the hypotenuse as "c", and the lengths of the other two sides as "a" and "b". So the equation is:

c² = a² + b²

To form an equation in x we can use a and b as variable, so we can substitute a = x and b = x-3, the equation becomes:

c² = x² + (x-3)²

To solve for c we can expand the second square and get:

c² = x² + x² - 6x + 9

Now we can add x² on both sides and get:

x² + x² = 6x - 9 + c²

We can combine like terms and get:

2x² = 6x - 9 + c²

Now we can move all the terms to one side and get:

2x² - 6x + 9 - c² = 0

Now we have an equation of a quadratic form with variable x, which can be solved using the quadratic formula.

x² = (6/2)x - 9/2 = 3x - 9/2 = 3x - 4.5 = 0

It can also be factored to get the solutions x = 4.5, x = 0

So the solutions to this equation are x = 4.5, x = 0.

(a) The lower bound of the confidence interval is 7.917 minutes.

(b) The upper bound of the confidence interval is 8.483 minutes.

(c) The margin of error is 0.283 minutes.

(a) The point estimate of the population mean is 8.2 minutes.

To calculate the lower bound of the 90% confidence interval, we use the formula: lower bound = point estimate - z* (standard error), where z* is the z-score for the 90% confidence level, which is 1.645, and the standard error is equal to the population standard deviation divided by the square root of the sample size, i.e., 2.2 / sqrt(200) = 0.156.

Substituting these values in the formula, we get the lower bound as 8.2 - 1.645 * 0.156 = 7.917 minutes.

(b) To calculate the upper bound of the 90% confidence interval, we use the same formula: upper bound = point estimate + z* (standard error). Substituting the values, we get the upper bound as 8.2 + 1.645 * 0.156 = 8.483 minutes.

(c) The margin of error is the difference between the point estimate and the upper or lower bound of the confidence interval. In this case, it is equal to half the width of the confidence interval, which is (8.483 - 7.917) / 2 = 0.283 minutes.

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Answer:
0.00465

Step-by-step explanation:

you just have to divide 1.1625 by 250

No, 11 is not an irreducible element in the ring R = Z[√-5]. The element 11 can be factored into smaller non-unit elements in R.

To determine if 11 is irreducible, we need to check if there exist non-unit elements a and b in R such that 11 = ab. In the ring R, an element a + b√-5 is a unit if and only if its norm, N(a + b√-5), is equal to ±1. The norm of a + b√-5 is defined as N(a + b√-5) = (a + b√-5)(a - b√-5) = a^2 + 5b^2.

For 11 = ab, we can try different values of a and b to see if we can find a non-unit factorization. If we let a = 1 + √-5 and b = 11, then we have:

(1 + √-5)(1 - √-5) = 1^2 + 5(-1) = 1 - 5 = -4

Therefore, we have found a non-unit factorization of 11 in R. Hence, 11 is not irreducible in R.

Regarding the prime ideal <11> in R, it is not a prime ideal because R is not even an integral domain. An integral domain is a commutative ring with unity where there are no zero divisors. In R = Z[√-5], zero divisors exist.

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The shape of the distribution of women's running times for the marathon at the 2012 London Olympics appears to be positively skewed, with a longer tail on the right-hand side of the boxplot.

A boxplot, also known as a box-and-whisker plot, is a graphical representation of the distribution of a dataset. It displays key statistical measures such as the median, quartiles, and outliers. In this case, the boxplot is used to represent the distribution of women's running times for the marathon at the 2012 London Olympics.

The box in the boxplot represents the interquartile range (IQR), which contains the middle 50% of the data. The line inside the box represents the median, or the 50th percentile, which is the value that separates the lower 50% and the upper 50% of the data. The whiskers, represented by lines extending from the box, show the range of the data within 1.5 times the IQR. Any data points outside of this range are considered outliers and are represented by individual data points or circles on the plot.

Based on the boxplot, it can be observed that the median (50th percentile) is closer to the lower quartile (25th percentile), while the upper quartile (75th percentile) is farther away from the median. This indicates that the majority of women's running times are concentrated towards the lower end of the distribution, with fewer data points towards the higher end. The longer tail on the right-hand side of the boxplot, as evidenced by the whisker extending beyond the upper quartile and the presence of outliers, suggests that there are some women who took longer times to finish the marathon, resulting in a positively skewed distribution.

Therefore, based on the shape of the boxplot, it can be concluded that the distribution of women's running times for the marathon at the 2012 London Olympics is positively skewed, with a longer tail on the right-hand side of the plot.

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Answer:

Step-by-step explanation:

2 with and exponent of 5 is 10 then you have to multiply that answer by 6 which will give you the answer of 60

So your final answer is 60

hope this helped :)

Answer:

19/25

Step-by-step explanation:

7/10 + 6/100

= 70/100 + 6/100

= 76/100

= 19/25

So, the answer is 19/25

The value of n is found as the 6 for the given exponent equation.

For the stated equation.

36 Superscript 12 minus n Baseline = 6 Superscript 2 n

This can be written as;

36¹²⁻ⁿ = 6²ⁿ

Using the adding exponent rule.

36¹².36⁻ⁿ = 6²ⁿ

Simplifying.

(6)²ˣ¹².(6)⁻ⁿˣ² = 6²ⁿ

(6)²⁴.(6)⁻²ⁿ = 6²ⁿ

Thus,

As, the base of number is equal , equating the powers.

24 - 2n = 2n

4n = 24

n = 6

Thus, the value of n is found as the 6 for the given exponent equation.

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The correct question is-

If 36 Superscript 12 minus n Baseline = 6 Superscript 2 n, what is the value of n?

Answer:B

Step-by-step explanation:

edge 2023

Answer:

Row related frequencies

                 twins in family history        Twins not in           Total

                                                             family history

 Twins born   8 / 34  = 0.2353                26 / 34= 0.7647     1                                                          

single child born :125 / 966 = 0.1294    841 / 966 = 0.8706   1                  

Step-by-step explanation:

Given table         twins in                  twins not in

                           family history             family history      total

Twins born                      8                          26                    34

single child born            125                       841                    966

 Total                              133                      867                   1000

Row related frequencies

Given table       twins in family history     Twins not in     Total      

                                                                        family history  

Twins born     8 / 34  = 0.2353      26 / 34= 0.7647                  1                                              

single child born 125 / 966 = 0.1294    841 / 966 = 0.8706      1                

Total                            0.3647                                1.6353                  

Answer: For the first 5 months she pays a monthly$15 membership fee. Use the explicit rule f(n) = 15n + 10

Step-by-step explanation:

The table of the ordered pairs and the graph is plotted.

The general equation of a straight line is -

[y] = [m]x + [c]

where -

[m] → is slope of line which tells the unit rate of change of [y] with respect to [x].

[c] → is the y - intercept i.e. the point where the graph cuts the [y] axis.

We have Emma who pays $10 to join a gym. For the first 5 months she pays a monthly $15 membership fee. The explicit rule that represents the situation above is f(n) = 15n+10.

We have the function as -

f(n) = 15n+10.

We can create a table as follows -

[n]         f[n]

1            25

2           40

3           55

4           70

5           85

Refer to the graph of the function f(n) = 15n+10 attached.

Therefore, the table of the ordered pairs and the graph is plotted.

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The US Crime Commission needs to take a sample of at least 1068 crimes to estimate the proportion of crimes in which firearms are used with a margin of error of 3% and a 95% confidence level, assuming no preliminary estimate is available.

To calculate the minimum required sample size, we can use the formula:

\(n = [(Z^2) \times p \times q] / E^2\)

where:

n is the sample size

Z is the Z-score corresponding to the desired confidence level (95% in this case), which is 1.96

p is the estimated proportion of crimes in which firearms are used (since no preliminary estimate is available, we can use 0.5 as a conservative estimate to get the maximum sample size)

q is the complement of p, which is 1 - p

E is the margin of error, which is 3% or 0.03

Substituting the values, we get:

\(n = [(1.96^2) \times 0.5 \times 0.5] / 0.03^2\)

n = 1067.11

Rounding up to the nearest integer, the minimum required sample size is 1068. Therefore, the US Crime Commission needs to take a sample of at least 1068 crimes to estimate the proportion of crimes in which firearms are used with a margin of error of 3% and a 95% confidence level, assuming no preliminary estimate is available.

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Answer:

x=10

Step-by-step explanation:

first, distribute the negative

4x+2-3x-2=10

combine like terms

x=10

\( - 34 < - 2(4x - 1) \\ - 34 < - 8x + 2 \\ - 34 - 2 < - 8x \\ - 36 < - 8x \\ 36 > 8x \\ \frac{36}{8} > x \\ x < \frac{9}{2} \)

The slope of an equation passing through the points (x₁, y₁) and (x₂, y₂) is:

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

a) The slope of the line passing through (-2, 5) and (4, 0) is -5/6.

b) The slope of the line passing through (0, 3) and (-2, 4) is -1/2.

c) The slope of the line passing through (-3, 4) and (-5, 6) is -1.

d) The slope of the line passing through (5, 5) and (3, 1) is 2.

e) The slope of the line passing through (-2, -1) and (-3, 1) is -2.

f) The slope of the line passing through (-4, -3) and (4, 1) is 1/2.

g) The slope of the line passing through (2, -1) and (2, 5) is undefined.

h) The slope of the line passing through (0, 2) and (1, 7) is 5.

i) The slope of the line passing through (3, 3) and (-3, 0) is 1/2.

j) The slope of the line passing through (0, 0) and (3, 3) is 1.

k) The slope of the line passing through (-4, 2) and (4,2) is 0.

l) The slope of the line passing through (-3, 5) and (-2,0) is -5.

m) The slope of the line passing through (2, 2) and (-3,-3) is 1.

n) The slope of the line passing through (-8, 10) and (-5, 6) is -4/3.

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The graph representing the interaction between meditation. Scull’s prediction that engaging in both activities does not produce any more benefit than either activity alone was wrong.

The interaction between exercise and meditation is more pronounced, indicating that it is necessary to engage in both activities to achieve better grades. Students who meditate and exercise regularly received better grades than those who did not meditate or exercise at all. According to the table of means, students who exercised but did not meditate had a mean of 3.6, students who meditated but did not exercise had a mean of 3.5, students who did not meditate or exercise had a mean of 2.5, and students who meditated and exercised had a mean of 3.8.

The mean score for the group who exercised but did not meditate was lower than the mean score for the group who meditated but did not exercise. The mean score for the group that neither meditated nor exercised was the lowest, while the group that meditated and exercised had the highest mean score.

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Answer:

1/3

Step-by-step explanation:

perpendicular lines have slopes that are opposite reciprocals

Answer:

I have not solved

Step-by-step explanation:

I am human being sonI didn't answer

Mark me as brainliest

1. True: If Spain has an absolute productivity advantage in producing shoes, then it will have a lower opportunity cost for producing shoes than the rest of the world, allowing them to sell them at a lower price, which would encourage exporting.

2. True: The Ricardian Model explains how nations can gain by specializing in the production of goods that they are relatively more efficient in producing and then trading. It can be used to explain how workers in the same sector can be differently affected due to international trade. 3. Uncertain: The extent to which a specific or mobile factor of production benefits from trade (or trade liberalization) depends on several factors, and cannot be generalized.

4. False: Suppose that Home has a relative abundance of L compared with Foreign, then it means that K is scarce relative to L in Home. Thus, K owners in Home will benefit from free trade policies as it will lead to an increase in the demand for K.

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A. -234

B. -62

C. \(-72+151=79\)

D. \(612+144=756\)

The equation is solved and the solution is d = 0.676

Given data ,

Mrs. Boyler wrote an equation on the board. She asked the students to solve for the variable, d

Akeem solved it on his paper using the following steps.

Step 1 d over 0.13 equals 5.2

Step 2 the product of 0.13 and d over 0.13 equals 5.2 divided by 0.13

Step 3 d = 40

Let the equation be represented as A

Now , the value of A is

d / 0.13 = 5.2

Multiply by 0.13 on both sides , we get

d = 0.13 ( 5.2 )

d = 0.676

Now , on further simplification , we get

d = 0.676

Hence , the solution is 0.676

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The complete question is attached below :

Mrs. Boyler wrote an equation on the board. She asked the students to solve for the variable, d. Akeem solved it on his paper using the following steps. Step 1 d over 0.13 equals 5.2 Step 2 the product of 0.13 and d over 0.13 equals 5.2 divided by 0.13 Step 3 d = 40 Part A: Review Akeem's work for solving the equation. State the step in which the error occurred, and describe the error using your mathematics vocabulary. (8 points) Part B: Solve the equation, correcting any errors you may find. Show all the steps in your work.

The p-chart is used to monitor the proportion of defective items in a sample.

The upper and lower 3-sigma control chart limits for a defect rate, we need to use the p-chart formula.

The p-chart is used to monitor the proportion of defective items in a sample. Given a historical defect rate of 2.00% (0.02) and a sample size of 400, we can calculate the control limits.

The formula for the control limits is:

UCLp = p + 3 * sqrt((p * (1 - p)) / n)

LCLp = p - 3 * sqrt((p * (1 - p)) / n)

Substituting the values, we get:

UCLp = 0.02 + 3 * sqrt((0.02 * (1 - 0.02)) / 400) ≈ 0.0263

LCLp = 0.02 - 3 * sqrt((0.02 * (1 - 0.02)) / 400) ≈ 0.0137

The upper control limit (UCLp) is approximately 0.0263, and the lower control limit (LCLp) is approximately 0.0137.

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Answer:

the correct answer is C

4x=52

52 divided in 4 parts is 13, so each ticket is 13 dollars

Skylar need to invest $101.36, to the nearest ten dollars, for the value of the account to reach $300 in 17 years.

We know that the formula for the compound interest is:

\(A = P(1 + r/n)^{nt}\)

where, A = Accrued amount

P = Principal amount    

R = interest rate as a percent

r =  interest rate as a decimal

r = R/100

n = number of compounding periods

t = time in years

Here, A = $300

t = 17 years

R = 6.4%

First we convert R as a percent to r as a decimal

r = R/100

r = 6.4/100

r = 0.064 per year,

Using above formula of compound interest,

P = A / (1 + r/n)nt

P = 300.00 / (1 + 0.064/12)(12)(17)

P = 300.00 / (1 + 0.0053333333333333)(204)

P = $101.36

Therefore, $101.36 need to invest.

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There sould be 11 terms of the following convergent series must be summed to be sure that the remainder is less than 10⁻¹ in magnitude

We can use the formula for the remainder of a convergent series:

Rn = Sn - S

where Rn is the remainder after summing the first n terms, Sn is the sum of the first n terms, and S is the infinite sum. In this case, the series is

sum(k=0 to infinity) (3^k - 1) / k!

so the infinite sum is S = e³ - 1. We want to find n such that the remainder Rn is less than 0.1, or |Rn| < 0.1.

The remainder formula tells us that

|Rn| = |sum(k=n+1 to infinity) (3^k - 1) / k!| <= sum(k=n+1 to infinity) |(3^k - 1) / k!|

To make an estimate, we can use the fact that

|3^k / k!| <= 3^k / (k-1)!

So we can write

|Rn| <= sum(k=n+1 to infinity) (3^k / (k-1)!) = e³ / (n!)

We want this to be less than 0.1, so we solve for n:

e³ / (n!) < 0.1

n! > e³ / 0.1

n > ln(e³ / 0.1)

n > 10.026

Since we can't sum a fraction of a term, we need to sum at least n = 11 terms to be sure that the remainder is less than 0.1 in magnitude.

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