Two monomials are shown below. 450x²y5 3000x4y³ What is the least common multiple (LCM) of these monomials? O 2xy O 30xy 150x²y³ 9000x¹y5
A2xy
b30xy
c15x^2y^3
d9000x^4y^5​

Answer:

y = 3x - 2

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

here slope m = 3 , then

y = 3x + c ← is the partial equation

to find c substitute (3, 7 ) into the partial equation

7 = 3(3) + c = 9 + c ( subtract 9 from both sides )

- 2 = c

y = 3x - 2 ← equation of line

The probability that

a) all seven children are boys is 0.0078 or around 0.78%.

b) all the children not of the same sex is  0.9844 or roughly 98.44%.

c) at least four of the children are girls is 0.2734, or roughly 27.34%. 

a) The likelihood of a child being a boy is 0.5, and expecting that the sexual orientation of each child is free of the others, the likelihood of all seven children being boys is (0.5)[tex]^7[/tex], which is 0.0078 or around 0.78%.

b) The likelihood of all children not being of the same sex is the complement of the likelihood that they are all of the same sex. There are two cases to consider:

either all the children are boys or all the children are young ladies. We as of now calculated the likelihood of all the children being boys in portion which is 0.0078.

The likelihood of all the children being young ladies is additionally (0.5)[tex]^7[/tex], which is additionally 0.0078. In this manner, the likelihood of all the children being of the same sex is 0.0078 + 0.0078 = 0.0156, and the likelihood of all the children not being of the same sex is

1 - 0.0156 = 0.9844 or roughly 98.44%.

c) The likelihood of at slightest four of the children being young ladies can be calculated utilizing the binomial dispersion. The likelihood of a young lady is 0.5, and the likelihood of a boy is too 0.5.

The likelihood of getting at slightest four young ladies can be calculated by including the probabilities of getting precisely 4, 5, 6, or 7 young ladies. Utilizing the binomial equation, the likelihood of getting precisely k young ladies out of n children is given by:

P(k young ladies) = (n select k) * [tex](0.5)^k * (0.5)^(n-k)[/tex]

where (n select k) is the binomial coefficient, which speaks to the number of ways to select k things out of n things. Utilizing this equation, we are able to calculate the likelihood of getting at slightest four young ladies as takes after:

P(at slightest 4 girls) = P(4 young ladies) + P(5 young ladies) + P(6 young ladies) + P(7 young ladies)

= [tex](7 select 4) * (0.5)^4 * (0.5)^3 + (7 select 5) * (0.5)^5 * (0.5)^2 + (7 select 6) * (0.5)^6 * (0.5)^1 + (7 select 7) * (0.5)^7 * (0.5)^0[/tex]

= 0.2734

Hence, the likelihood of at slightest four of the children being young ladies is 0.2734, or roughly 27.34%. 

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The expression "-3x² + 3x²" simplifies to 0, so the given function can be rewritten as: f(x) = -x - 1.

To find the first derivative of f(x), we use the power rule and the constant multiple rule of differentiation: f'(x) = -1. The first derivative of f(x) is simply -1, which means that the slope of the tangent line to the graph of f(x) is constant and equal to -1 for all values of x.

To find the second derivative of f(x), we differentiate the first derivative:

f''(x) = 0. The second derivative of f(x) is 0, which means that the graph of f(x) is a straight line with a constant slope of -1, and it has no curvature or inflection points.

When we graph f(x), f'(x), and f''(x) using a graphing calculator or software, we can see that the graph of f(x) is a straight line with a negative slope of -1, as expected. The graph of f'(x) is a horizontal line at y = -1, which confirms that the slope of f(x) is constant. The graph of f''(x) is a horizontal line at y = 0, which confirms that f(x) has no curvature or inflection points.

The analysis of the first and second derivatives of f(x) reveals that the function is a straight line with a constant negative slope, and it has no curvature or inflection points.

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As h approaches 0, the limit becomes: e(²+0) = e²
The answer is D, e².

The given expression is lim(h→0) (e(2+h) - e²)/h. To find the limit, we can apply L'Hopital's Rule since we have an indeterminate form of the type 0/0. L'Hopital's Rule states that if lim(f(x)/g(x)) as x→a is indeterminate, then it is equal to lim(f'(x)/g'(x)) as x→a, provided the limit exists.

Here, f(h) = e^(²+h) - e^² and g(h) = h. Let's find their derivatives:

f'(h) = d(e(²+h) - e^²)/dh = e^(²+h)
g'(h) = dh/dh = 1

Now, applying L'Hopital's Rule:

lim(h→0) (e(²+h) - e²)/h = lim(h→0) (e²+h))/1

As h approaches 0, the limit becomes:

e(²+0) = e²

So, the answer is D, e².

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Using the three-point Gauss quadrature formula, we get I ≈ 0.0817.

(a) Using the composite trapezoidal rule with n = 1, 2, and 4, we get:
For n = 1: I ≈ (b-a) / 2 [f(a) + f(b)] = 1/2 [0 + 1/4] = 1/8
For n = 2: I ≈ (b-a) / 4 [f(a) + 2f(a+h) + f(b)] = 1/4 [0 + 1/8 + 1/4] = 3/32
For n = 4: I ≈ (b-a) / 8 [f(a) + 2f(a+h) + 2f(a+2h) + 2f(a+3h) + f(b)] = 1/8 [0 + 1/8 + 1/2 + 1/2 + 1/4] = 11/64

(b) Using Romberg extrapolation twice, we get:
R(1,1) = 1/8, R(2,1) = 3/32, R(4,1) = 11/64
R(2,2) = [4R(2,1) - R(1,1)] / [4 - 1] = 7/64
R(4,2) = [4R(4,1) - R(2,1)] / [4 - 1] = 59/256
So, the more accurate estimate of the integral using Romberg extrapolation twice is R(4,2) = 59/256.

(c) Using the two-point Gauss quadrature formula, we get:
I ≈ (b-a) / 2 [f((a+b)/2 - (b-a)/(2sqrt(3))) + f((a+b)/2 + (b-a)/(2sqrt(3)))]
= 1/2 [0.0728 + 0.1456] = 0.1092

(d) Using the three-point Gauss quadrature formula, we get:
I ≈ (b-a) / 2 [5/9 f((a+b)/2 - (b-a)/(2sqrt(15))) + 8/9 f((a+b)/2) + 5/9 f((a+b)/2 + (b-a)/(2sqrt(15)))]
= 1/2 [0.0146 + 0.1343 + 0.0146] = 0.0817

Therefore, using the composite trapezoidal rule, we get I ≈ 11/64. Using Romberg extrapolation twice, we get a more accurate estimate of I ≈ 59/256. Using the two-point Gauss quadrature formula, we get I ≈ 0.1092. Using the three-point Gauss quadrature formula, we get I ≈ 0.0817.

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The mean and standard deviation of the Poisson distribution are μ = 9.0 and σ = 3.0.

In a Poisson distribution, the mean and standard deviation are equal to the parameter λ. Therefore, in this case, μ = σ = λ = 9.0.

The formula for the mean and standard deviation of a Poisson distribution are:

Mean (μ) = λ

Standard Deviation (σ) = √λ

Substituting λ = 9.0, we get:

μ = 9.0

σ = √9.0 = 3.0

Therefore, the mean and standard deviation of the Poisson distribution are μ = 9.0 and σ = 3.0.

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When using regression analysis to estimate the index model for a stock, the slope of the regression line is an estimate of the β of the asset.

The beta (β) of an asset measures the sensitivity of the asset's returns to changes in the market as a whole. A beta of 1 indicates that the asset's returns move in lockstep with the market, while a beta greater than 1 indicates that the asset's returns are more volatile than the market and a beta less than 1 indicates that the asset's returns are less volatile than the market.

On the other hand, α (alpha) is a measure of the excess return of an asset relative to its expected return, given its beta. α is not estimated by the slope of the regression line, but rather by the intercept of the regression line.

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Answer:

16.25 miles

Step-by-step explanation:

We Know

Mai's family travels in a car at a constant speed of 65 miles per hour.

How far do they travel in 25 minutes?

25 minutes = 1/4 of an hour

We Take

65 / 4 = 16.25 miles

So, they travel 16.25 miles in 25 minutes.

The particle's position at time t=0 is 3/2.

That velocity-time is the derivative of the position function, so I thought I could find the anti-derivative of v(t)

and used the given position to solve for the integration constant and then would have a formula of the position which would allow for me to solve for the t=0.

x(t)=∫sin(2t)dt

=−1/2 * cos(2t)+C

So,

time, t = π/2 ; x = 4

4 = −1/2cos(2[π/2])+C

4 = -1/2 * cos π + C

4 =  -1/2 * (-1) + C

C= 4−1/2

= 5/2

Now, put C = 5/2 to find particle position at time, t=0:

x(t) = −1/2 * cos(2t)+C

= −1/2 * cos(2(0)) + 5/2

= -1/2 + 5/2

= 3/2

Hence,  the particle's position at time t=0 is 3/2.

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The number of hose is multiplied by 8 to get the number of gallons

From the question, we have the following parameters that can be used in our computation:

The table of values

The constant, k of the ratio is calculated as

k = y/x

So, we have

k = 80/10 = 40/5.....

Evaluate

k = 8

This means that the relationship is a multiplicative relationship because the number of hose is multiplied by 8 to get the number of gallons

Hence, the true statement is (c)

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This value is approximately equal to -0.726. To use the sequence of partial sums to prove convergence, we need to find the limit of the sequence of partial sums. The partial sum for the first n terms of the series is:

Sn = Ση=4n 5/n2-31

We want to show that this sequence of partial sums converges to some limit L. To do this, we can use the fact that the series is absolutely convergent. This means that the series of absolute values converges, which implies that the series itself converges. We can see that the series of absolute values is:

Ση=4n |5/n2-31|

Since all terms are positive, we can drop the absolute value signs:

Ση=4n 5/n2-31

Now, we can use the comparison test to show that this series converges. We know that:

5/n2-31 < 5/n2

Therefore, we can compare our series to the series:

Ση=1∞ 5/n2

which we know converges by the p-test. Since the terms of our series are smaller than the terms of the convergent series, our series must also converge.

Now that we have shown that the series converges, we can find its limit L by taking the limit of the sequence of partial sums. That is:

lim n→∞ Ση=4n 5/n2-31 = L

We can use the fact that the series is absolutely convergent to rearrange the terms of the series:

Ση=4n 5/n2-31 = Ση=1n 5/η2-31 - Ση=1∞ 5/η2-31

The second series on the right-hand side is a convergent series, so it must have a finite sum. Therefore, as n approaches infinity, the sum of the first series on the right-hand side approaches the sum of the entire series:

lim n→∞ Ση=1n 5/η2-31 = L + Ση=1∞ 5/η2-31

Solving for L, we get:

L = lim n→∞ Ση=1n 5/η2-31 - Ση=1∞ 5/η2-31

Since we know that the second series on the right-hand side has a finite sum, we can evaluate it:

Ση=1∞ 5/η2-31 = 5/1-31 + 5/4-31 + 5/9-31 + ...

This is a convergent p-series with p=2, so we can evaluate it using the formula:

Ση=1∞ 1/η2 = π2/6

Substituting this value into our expression for L, we get:

L = lim n→∞ Ση=1n 5/η2-31 - π2/6

We can evaluate the limit using the integral test:

∫1∞ 5/x2-31 dx = lim n→∞ Ση=1n 5/η2-31

This integral evaluates to:

lim t→∞ 5/sqrt(31)(arctan(sqrt(31)/t) - arctan(sqrt(31)))

= 5/sqrt(31) * π/2

Therefore, our final answer is:

L = 5/sqrt(31) * π/2 - π2/6

Note that this value is approximately equal to -0.726.

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The 90% confidence interval for the proportion of homeless persons who are veterans is (0.110, 0.240) based on a sample of 200.

To build a 90% certainty span for the extent of destitute people who are veterans, we really want to compute the standard blunder of the example extent and afterward duplicate it by the basic worth of the standard typical dispersion relating to the picked certainty level (1.65 for 90% certainty). Involving the recipe for the standard blunder of an example extent, we get a standard mistake of 0.032. Then, at that point, we work out the lower and upper limits of the certainty span by deducting and adding the result of the standard mistake and the basic worth to the example extent, individually. Consequently, the 90% certainty span for the extent of destitute people who are veterans is (0.110, 0.240), implying that we are 90% sure that the genuine extent of destitute people who are veterans falls inside this reach.

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The probability of having 2 trucks with no violations, 2 with 1 violation, and 3 with 2 or more violations when 7 trucks are inspected are approximately 0.041 (rounded to three decimal places).

To find the probability of this specific situation, we can use the multinomial probability formula:
P(X) = (n! / (x1! × x2! × ... × xn!)) × (p1^x1) × (p2^x2) × ... × (pn^xn)
In this case:
- n = 7 (total number of trailer trucks)
- x1 = 2 (trucks with no violations)
- x2 = 2 (trucks with 1 violation)
- x3 = 3 (trucks with 2 or more violations)
- p1 = 0.1 (probability of no violations)
- p2 = 0.3 (probability of 1 violation)
- p3 = 0.6 (probability of 2 or more violations)
Now we plug in the values:
P(X) = (7! / (2! × 2! × 3!)) × (0.1^2) × (0.3²) × (0.6³)
P(X) = (5040 / (2 × 2 × 6)) × (0.01) × (0.09) × (0.216)
P(X) = (210) × (0.01) × (0.09) × (0.216)
P(X) ≈ 0.040788

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The  function  [tex]Q(t)=8^t^/^3[/tex]  is exponential and values of a is 1 and b is  [tex]8^(^1^/^3^)[/tex]

Yes, the function [tex]Q(t)=8^t^/^3[/tex] is exponential.

We can write it in the form [tex]f(t) = ab^t[/tex],

where: a = Q(0) = [tex]8^(^0^/^3^)[/tex]

= 8⁰

= 1

b = [tex]8^(^1^/^3^)[/tex]

Therefore, the function Q(t) in the form of [tex]f(t) = ab^t[/tex],is:

f(t) = 1 × [tex]8^(^1^/^3^)^t[/tex]

[tex]f(t) = 8^(^t^/^3^)[/tex]

So, values a = 1 and b = [tex]8^(^1^/^3^)[/tex]

Hence, the given function  [tex]Q(t)=8^t^/^3[/tex]  is exponential and values of a is 1 and b is  [tex]8^(^1^/^3^)[/tex]

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Is the function Q(t) = 8^(t/3) exponential?

If yes, write the function in the form of f(t)=ab^t and enter the values of a and b

The projection of u onto v is approximately 5.244i + 4.195j, and the vector component of u orthogonal to v is approximately -2.244i + 2.805j.

To find the projection of u onto v and the vector component of u orthogonal to v, we'll need to use the formulas for projection and orthogonal components. Let's start with part (a):

(a) To find the projection of u onto v, we'll use the formula:

proj(u onto v) = (u • v / ||v||²) * v

where u = 3i + 7j, v = 5i + 4j, and "•" represents the dot product.

First, let's find the dot product of u and v:
u • v = (3 * 5) + (7 * 4) = 15 + 28 = 43

Next, find the squared magnitude of v:
||v||² = (5² + 4²) = 25 + 16 = 41

Now, divide the dot product by the squared magnitude:
43 / 41 ≈ 1.0488

Finally, multiply this value by the vector v:
proj(u onto v) ≈ 1.0488 * (5i + 4j) ≈ 5.244i + 4.195j

Now let's move to part (b):

(b) To find the vector component of u orthogonal to v, we'll use the formula:

u_orthogonal = u - proj(u onto v)

We've already calculated proj(u onto v) as 5.244i + 4.195j. Now we just need to subtract this from the original vector u:

u_orthogonal = (3i + 7j) - (5.244i + 4.195j) ≈ (-2.244i) + 2.805j

So,The projection of u onto v is approximately 5.244i + 4.195j, and the vector component of u orthogonal to v is approximately -2.244i + 2.805j.

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Based on the mentioned informations and provided values, the value of  the function g'(3a)  is calculate to be equal to 11/21.

We can start by applying the chain rule to find the derivative of g(x) with respect to x:

g'(x) = (1/7) f'(x/3) (1/3)

Note that the factor of 1/3 comes from the chain rule, since we are differentiating with respect to x but the argument of f is x/3.

Next, we can substitute x = 3a to find g'(3a):

g'(3a) = (1/7) f'(3a/3) (1/3)

= (1/7) f'(a) (1/3)

= (1/7) (11) (1/3)

= 11/21

Therefore, g'(3a) = 11/21.

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On solving the query we can say that Answer: 33%, rounded to the function closest full number. D. 33% is the answer that is closest to the real one.

Mathematics is concerned with numbers and their variations, equations and related structures, shapes and their places, and possible placements for them. The relationship between a collection of inputs, each of which has an associated output, is referred to as a "function". An relationship between inputs and outputs, where each input yields a single, distinct output, is called a function. Each function has a domain and a codomain, often known as a scope. The letter f is frequently used to represent functions (x). X is the input. The four main types of functions that are offered are on functions, one-to-one functions, many-to-one functions, within functions, and on functions.

The basketball squad has a total of 15 players, which is equal to 3 + 5 + 3 + 4.

There are five sophomores.

We may use the following formula to get the team's proportion of sophomores:

(Part/Whole) x 100 equals %

In this instance, the "part" is the quantity of sophomores, which is 5, and the "whole" is the overall player count, which is 15. Thus:

% = (5/15) multiplied by 100 percent equals 33.33

Answer: 33%, rounded to the closest full number. D. 33% is the answer that is closest to the real one.

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The implications of this analysis for the advertising agency are the advertising agency should strive to create commercials that appeal to both men and women, taking into account the differences in their television viewing habits. The option (c) is correct.

To test the manager's claim that there is a significant gender difference in television viewing, we need to conduct a t-test for independent means. The null hypothesis is that there is no significant difference in the amount of television watched between men and women, while the alternative hypothesis is that there is a significant difference.

The t-test for independent means gives us a t-value of -2.44, which is significant at the .05 level. This means that we can reject the null hypothesis and conclude that there is a significant gender difference in television viewing.

In terms of implications for the advertising agency, it would be wise to consider the differences in television viewing habits between men and women when creating commercials. Based on the data, women watch more television on average than men, so the agency may want to tailor their commercials more towards the interests and needs of women. This does not mean that men should be ignored entirely, as they still make up a significant portion of the viewing audience.

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Rounded to 3 decimal places, the variance estimates are: var1 = 26.476 var2 = 27.524 ,  Rounded to 3 decimal places, the variance estimates are: var1 = 33.538 ,var2 = 35.333 , Rounded to 3 decimal places, the variance estimates are: var1 = 27.381, var2 = 30.833 , The data will tend to be more tightly clustered around it, resulting in a smaller variance.

To compute the variance estimates for Data Set A using the definitional formula provided:

a) Using the sample mean for Data Set A:

First, we need to calculate the sample mean for each column of the data set:

x1 = (23 + 13 + 13 + 7 + 9 + 19 + 11 + 19 + 15 + 14 + 17 + 21 + 21 + 17) / 14

  = 15.14

x2 = (19 + 7 + 19 + 19 + 14 + 21 + 17) / 7

  = 16.43

Using these sample means, we can calculate the variance of each column using the definitional formula:

var1 = [tex][(23-15.14)^2 + (13-15.14)^2 + ... + (21-15.14)^2] / 13[/tex]

    = 26.48

var2 =[tex][(19-16.43)^2 + (7-16.43)^2 + ... + (17-16.43)^2] / 6[/tex]

    = 27.52

Rounded to 3 decimal places, the variance estimates are:

var1 = 26.476

var2 = 27.524

b) Using a mean score of 15:

Using a mean score of 15, we can calculate the variance of each column using the same formula as in part (a), but with the mean score of 15 substituted for the sample mean:

var1 = [tex][(23-15)^2 + (13-15)^2 + ... + (21-15)^2] / 13[/tex]

    = 33.54

var2 = [tex][(19-15)^2 + (7-15)^2 + ... + (17-15)^2] / 6[/tex]

    = 35.33

Rounded to 3 decimal places, the variance estimates are:

var1 = 33.538

var2 = 35.333

c) Using a mean score of 16:

Using a mean score of 16, we can calculate the variance of each column using the same formula as in part (b), but with the mean score of 16 substituted for 15:

var1 = [tex][(23-16)^2 + (13-16)^2 + ... + (21-16)^2] / 13[/tex]

    = 27.38

var2 = [tex][(19-16)^2 + (7-16)^2 + ... + (17-16)^2] / 6[/tex]

    = 30.83

Rounded to 3 decimal places, the variance estimates are:

var1 = 27.381

var2 = 30.833

d) Conclusions:

The variance estimates are sensitive to the choice of mean score used in the calculations. In general, the variance estimates will be larger when a mean score that is lower than the sample mean is used, and smaller when a mean score that is higher than the sample mean is used. This is because the variance measures the spread of the data around the mean, and if the mean is shifted higher, the data will tend to be more tightly clustered around it, resulting in a smaller variance. Similarly, if the mean is shifted lower, the data will tend to be more spread out, resulting in a larger variance.

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To find the expected value of a random variable, we multiply each possible value of the variable by its probability and then add up the products.
So, the expected value of X can be calculated as:
E(X) = (0)(0.5) + (1)(0.2) + (2)(0.3)
    = 0 + 0.2 + 0.6
    = 0.8
Therefore, the answer is c. 0.80.

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1. The answer for the first integral is: [tex]-2/x^2 + 7ln|x| + C[/tex]
2. The answer for the second integral is: [tex](1/7)e^{7x} + C[/tex]

Let's solve each of them:
1) [tex]\int (4/x^3 + 7/x) dx[/tex]
Rewrite the integral with each term separately:
[tex]\int (4/x^3) dx + ∫(7/x) dx[/tex]
Integrate each term:
[tex]-2/x^2 + 7ln|x| + C[/tex]
So, the answer for the first integral is: [tex]-2/x^2 + 7ln|x| + C[/tex]
2) [tex]\int e^7x dx[/tex]
Use the integration rule for exponential functions:
[tex]\int e^{ ax}  dx = (1/a)e^{ax} + C[/tex]
In this case, a = 7.
Apply the rule:
[tex](1/7)e^{7x} + C[/tex].

The integration rule for exponential functions is:

∫ [tex]e^x dx = e^x + C[/tex]

where C is the constant of integration.

This rule can be used to integrate any function of the form[tex]f(x) = e^x[/tex], where e is the mathematical constant approximately equal to 2.71828.

To use this rule, we simply replace f(x) with [tex]e^x[/tex] in the integral and then apply the rule.

For example:

∫[tex]e^{3x} dx = (1/3)e^{3x} + C[/tex]

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The probability that exactly 4 out of 6 people chosen at random favor the highway reconstruction project is 0.0154, or about 1.54%.

To solve this problem, we can use the binomial distribution formula:
P(X = k) = (n choose k) × [tex]p^k[/tex] × [tex](1-p)^{(n-k)}[/tex]
Where:
- X is the number of people who favor the highway reconstruction project (in this case, k = 4)
- n is the total number of people chosen at random (in this case, n = 6)
- p is the probability of an individual favoring the highway reconstruction project (in this case, p = 0.20)
Plugging in the values, we get:
P(X = 4) = (6 choose 4) × [tex]0.20^4[/tex] × [tex](1-0.20)^{(6-4)}[/tex]
P(X = 4) = (15) × 0.0016 × 0.64
P(X = 4) = 0.0154

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There are no specific values of x for which the rate of change is minimum.

function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions are ubiquitous in mathematics and are essential for formulating physical relationships in the sciences.

To find the value(s) of x for which the rate of change of the function f(x) = 3.75 - 5.7 + 50 - 7 is minimum, first, we need to simplify the function:

f(x) = 3.75 - 5.7 + 50 - 7
f(x) = -1.95 + 43
f(x) = 41.05

Since the function f(x) is a constant function, its rate of change is always 0, and it does not have a minimum or maximum value.

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The arithmetic mean of this data set is: 69.88
The median of this data set is: 68.2
The mode of this data set is: 84.7

To calculate the mean, median, and mode of the given data set, follow these steps:

1. Arrange the data set in ascending order: 49.5, 56.6, 57.1, 63.8, 68.2, 69.8, 73.4, 84.7, 84.7, 84.7

2. Calculate the mean by adding all the numbers and dividing by the total count:
(49.5+56.6+57.1+63.8+68.2+69.8+73.4+84.7+84.7+84.7) / 10 = 692.5 / 10 = 69.25
Mean = 69.25

3. Calculate the median by finding the middle value(s) of the ordered data set. In this case, there are 10 numbers, so we will take the average of the two middle values (5th and 6th):
(68.2 + 69.8) / 2 = 138 / 2 = 69
Median = 69

4. Calculate the mode by identifying the number(s) that appear most frequently. In this case, 84.7 appears three times:
Mode = 84.7

Your answer: The mean of the data set is 69.25, the median is 69, and the mode is 84.7.

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Step-by-step explanation:

Law of Cosines

c^2 = a^2 + b^2  - 2 ab cosΦ

for this example   c = 8     a = 16     b = 17    Φ= Angle C

  you are trying to solve for Angle C

8^2 = 16^2 + 17^2 - 2 (16)(17) cos C

-481 = -2 (16)(17) cos C

.88419 = cos C

arccos ( .884190 ) = C = 27.8 degrees

Answer:

Unit Conversion:

l≈3.35m

w≈0.2m

Solution

P=2(l+w)=2·(3.35+0.2)=7.112m

P=280

Answer: 38 in

Step-by-step explanation:

8+8+(11 x 12) + (11x12)

16 + 132 + 132

280

To answer this question, we will use the concepts of mean number and standard deviation in the context of a binomial distribution. In this case, the number of trials (n) is 65, and the probability of success (learning to push the lever) is 0.70.

(a) Find the mean number of rats that will learn to push the lever:

Mean (µ) = n * p
Mean (µ) = 65 * 0.70
Mean (µ) = 45.5

So, on average, 45.5 rats will learn to push the lever.

(b) Find the standard deviation of the number of rats that will learn to push the lever:

Standard Deviation (σ) = √(n * p * q)
Where q is the probability of failure (1 - p)

In this case, q = 1 - 0.70 = 0.30

Standard Deviation (σ) = √(65 * 0.70 * 0.30)
Standard Deviation (σ) = √(13.65)
Standard Deviation (σ) ≈ 3.69

The standard deviation of the number of rats that will learn to push the lever is approximately 3.69 (rounded to two decimal places).

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The correct answer is C. The distribution is negatively skewed.

In a boxplot, the box represents the interquartile range (IQR), which contains the middle 50% of the data. The median (m) is represented by a vertical line inside the box. The whiskers extend from the box to the smallest and largest observations within 1.5 times the IQR of the box. Any observations beyond the whiskers are considered outliers.

In this boxplot, the median (m) is closer to the bottom whisker than to the top whisker, which suggests that the distribution is negatively skewed. Additionally, the box appears to be longer on the left side than on the right side, which further supports the conclusion that the distribution is negatively skewed. Therefore, the correct answer is C. The distribution is negatively skewed.

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According to the information, the order of the ribbons from the lowest to the highest would be: 1.73, 2.23, 3.13, 3.46

To find the root of a number we can use different methods such as:

According to the above information, the results of the roots would be:

1.73 = [tex]\sqrt{3}[/tex]

2.23 = [tex]\sqrt{5}[/tex]

3.13 = π

3.46 = [tex]2\sqrt{3}[/tex]

So the order from lowest to highest of the ribbons according to their value would be:

1.73, 2.23, 3.13, 3.46

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