five and four hundred three thousandths in standard form

The population proportion who participated in voting (63%) is higher than the sample proportion from this survey (58.91%).

To determine whether the population proportion who participated in voting or the sample proportion from the survey is higher, we need to compare the percentages.

The population proportion who participated in voting is given as 63% of all registered voters.

This means that out of every 100 registered voters, 63 participated in voting.

In the survey of retired voters, 162 out of 275 participants voted. To calculate the sample proportion, we divide the number of retired voters who participated (162) by the total number of retired voters in the sample (275) and multiply by 100 to get a percentage.

Sample proportion = (162 / 275) \(\times\) 100 ≈ 58.91%, .

Comparing the population proportion (63%) with the sample proportion (58.91%), we can see that the population proportion who participated in voting (63%) is higher than the sample proportion from this survey (58.91%).

Therefore, based on the given data, the population proportion who participated in voting is higher than the sample proportion from this survey.

It's important to note that the sample proportion is an estimate based on the surveyed retired voters and may not perfectly represent the entire population of registered voters.

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Answer:

y=x-4

Step-by-step explanation:

The correct answer is c. both A and B are correct. When one increases the confidence level from 0.90 to 0.95, the margin of error will increase because a higher level of confidence requires a wider interval to capture the population parameter.

However, the resulting confidence interval will also capture the population parameter more often because the higher confidence level provides a greater level of certainty that the interval contains the true population parameter.

When one increases the confidence level (1-α), say from 0.90 to 0.95, both A and B are correct.

Explanation:
a. The margin of error will increase because increasing the confidence level means we are more certain that the population parameter lies within the confidence interval. To achieve this, the interval has to be wider, which results in a larger margin of error.

b. The resulting confidence interval will capture the population parameter more often because a higher confidence level indicates a higher probability that the interval contains the true population parameter. In this case, going from a 0.90 to 0.95 confidence level means that we would expect the interval to capture the true population parameter 95% of the time, as opposed to 90% of the time.

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The value for given differential equation is ysoln1 = 3/10 + 1/2 cos t + 1/5 e-t (3 sin t + 2 cos t) + 1/5 e-t cos t, ysoln2 = 2 e-2t + 4 e3t + 2t e3t and ysoln3 = 2 t e t - 2 e t + e-2t.

The Laplace transform of the differential equation is :

D3 Y (s) + 2 D2 Y (s) + D Y (s) + 2 Y (s) = 5 + 4sin t

We know that,

L(Dn y(t))/dt^n = sn (L(y)) - sn-1 (y(0)) - sn-2 (y'(0)) - ... - sy(n-2) (0) - y(n-1) (0)

Putting n=0, 1, 2 in above formula, we get,

L(y) = Y (s), L(y') = sY (s) - y(0), L(y'')

                         = s2 Y (s) - s y(0) - y'(0)

                         = s2 Y (s) - 2s - 3

Substituting these values in the differential equation, we get :

s3 Y (s) - 3s2 - 3s + s Y (s) - 2 Y (s) = 5 + 4 L(sin t)

Taking Laplace transform of the differential equation, we get :

Y (s) = 5 s3 + s2 - 3s - 4s (s3 + 2s2 + s + 2

Using partial fraction, we get :

Y (s) = 1/2 s + 3/10 + 5/10 s + 7/10 s2 + 7/10 s + 1/5 (1/ (s2 + 2s + 1)) + (2 s - 1)/ (s2 + 2s + 1)

Taking inverse Laplace transform, we get :

ysoln1 = 3/10 + 1/2 cos t + 1/5 e-t (3 sin t + 2 cos t) + 1/5 e-t cos t

2. The Laplace transform of the differential equation is :

D2 Y (s) + 5 D Y (s) + 6 Y (s) = 5 + 3 e3t

We know that

L(Dn y(t))/dt^n = sn (L(y)) - sn-1 (y(0)) - sn-2 (y'(0)) - ... - sy(n-2) (0) - y(n-1) (0)

Putting n=0, 1 in above formula, we get,

L(y) = Y (s), L(y') = sY (s) - y(0)

Substituting these values in the differential equation, we get:s2 Y (s) - 5s - 6 + s Y (s) = 5 + 3 / (s - 3)

Taking Laplace transform of the differential equation,

we get :

Y (s) = 8 / (s - 3) + 1 / (s + 2) + 1 / ((s - 3)2)

Using partial fraction, we get :

Y (s) = 4 / (s - 3) + 2 / (s + 2) + 1 / (s - 3)2

Taking inverse Laplace transform, we get :

ysoln2 = 2 e-2t + 4 e3t + 2t e3t

3. The Laplace transform of the differential equation is :

D2 Y (s) + 6 D Y (s) + 4 Y (s) = 6 e x + 4t2

We know that

L(Dn y(t))/dt^n = sn (L(y)) - sn-1 (y(0)) - sn-2 (y'(0)) - ... - sy(n-2) (0) - y(n-1) (0)

Putting n=0, 1 in above formula, we get,

L(y) = Y (s), L(y') = sY (s) - y(0)

Substituting these values in the differential equation, we get :

s2 Y (s) - s y(0) - y'(0) + 6 s Y (s) - 6 y(0) + 4 Y (s) = 6 / (s - 1)2

Taking Laplace transform of the differential equation, we get :

Y (s) = 6 / (s - 1)2 (s2 + 6s + 4)

Using partial fraction, we get :

Y (s) = 2 / (s - 1)2 - 2 / (s - 1) + 1 / (s + 2)

Taking inverse Laplace transform, we get :

ysoln3 = 2 t e t - 2 e t + e-2t  

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t distributions tend to be flatter and more spread out than the normal distribution.

This is due to the fact that in the formula, the denominator is s rather than σ.

The distribution of the sample mean is normal if some samples are taken from a normal population with known variance. However, if the population variance is unknown, the distribution is not normal but Student-t with long tails. This means that sample means tend to be extreme when the population variance is unknown. Using the normal distribution instead of the t distribution to test the hypothesis increases the chance of error.

Note that there is a different t-distribution for each sample size. That is, the class of distributions. When talking about a particular t-distribution, we need to specify the degrees of freedom. The degrees of freedom for this t-statistic is given by the sample standard deviation s in the denominator of Equation 1. The spread is larger than the standard normal distribution. This is because the denominator of equation (1) is s, not σ. Because s is a random variable that changes from sample to sample, t becomes more volatile and more spread out.

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Answer:

1 11/20

Step-by-step explanation:

3 3/4 - 2 1/5

(3-2) + (3/4 - 1/5)

1 + 11/20

1 11/20 or 31/20 or 1.55

The pond will be filled at 01:30 PM.

If Brittany started filling the pond at 9:30 A.M. and it takes him 1 hour to fill 1,600 gallons, then to fill 3,200 gallons, it will take him 3,200 / 1,600 = 2 hours.

The capacity of pond= 3600 gallon.

The Burt is started at 9:30 AM  and filled 1600 gallons till 11:30 AM.

So, in 2 hours the Burt had filled 1600 gallons.

So, rate of filling = 1600/2=800 gallon/hour

The remaining volume of the pond = 3200-1600 = 1600 gallons

Time to fill the remaining volume = 1600/800=2 hour.

So, it will take 2 hours after 11:30 AM to fill the pond completely.

Hence, the pond will be filled at 01:30 PM.

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The volume of the solid that results when the region enclosed by the curves is revolved about the axis x = -1 is approximately 26.704 units cubed.

To find the volume of the solid that results when the region enclosed by the curves is revolved about the given axis, we use the method of cylindrical shells. The formula for the volume of a cylindrical shell is:

V = 2πrhΔx

where r is the distance from the axis of rotation to the shell, h is the height of the shell, and Δx is the thickness of the shell.

In this case, the axis of rotation is x = -1. We need to find the limits of integration for x and y in order to set up the integral for each problem.

The curves are \(y = x^2 - 1, y = -x,\) and x = -1. Solving for x in terms of y for each curve, we get:

\(x =± \sqrt{(y+1} )\) for y in [-1, 1]

x = -y for y in [-1, 0]

The limits of integration for y are -1 to 0 for the segment of the curve y = -x, and -1 to 1 for the segment of the curve \(y = x^2 - 1\). The limits of integration for x are -1 to √(y+1) for the upper half of the parabola, and -√(y+1) to -1 for the lower half.

The integral for the volume is:

V = 2π ∫[-1,0] ∫[-1,0] (-x - (-1))y dx dy + 2π ∫[-1,1] ∫[-√(y+1),√(y+1)] (√(y+1) - (-√(y+1)))y dx dy

Simplifying and evaluating this integral, we get V = 26.704.

Therefore, the volume of the solid that results when the region enclosed by the curves is revolved about the axis x = -1 is approximately 26.704 units cubed.

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We are given the following

we want to express m in terms of the other variables. So, what we will do is apply math operations on both sides, so we "isolate" the m variable on one side of the equation.

So, we begin by factoring the m on the right side, so we get

Now, we divide both sides by (a+3x). So we get

which is our final answer.

Answer:

-2y + 30

Step-by-step explanation:

3(y+5)-5(y-3)

Distribute

3y+15 -5y +15

Combine like terms

-2y + 30

Answer:

60 ft

Step-by-step explanation:

a² + b² = c²

20² + 15² = c²

c = √625

c = 25

total distance = 20 ft + 15 ft + 25 ft = 60 ft

The simplification of the expression 4a + 3b - a + 3b + 6 is 3(a + 2b + 2)

Simplify 4a + 3b - a + 3b + 6

collect like terms

= 4a - a + 3b + 3b + 6

= 3a + 6b + 6

factorize

= 3(a + 2b + 2)

side a = 3x - 5

side b = 2x - 1

side c = x + 1

Perimeter = 31 cm

The perimeter of a triangle = side a + side b + side c

31 = (3x - 5) + (2x - 1) + (x + 1)

31 = 3x - 5 + 2x - 1 + x + 1

31 = 6x - 5

Add 5 to both sides

31 + 5 = 6x

36 = 6x

divide both sides by 6

x = 36/6

x = 6

Therefore, the value of x from the perimeter of the triangle is 6

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Answer:179 cm3

there you go :)

Step-by-step explanation:

200 x 0.87 = 174

She answered 174 questions correctly.

Answer:

Chloe answered 174 problems correctly.

Explanation:

To find percentage, you would multiply the percentages decimal to the total.

0.87x200=174

Answer:

The significance test that would be used to investigate whether there is convincing evidence at 0.05 level of significance that a greater percent of peaches in the ice cream is associated with an increase in the density of the ice cream is option C;

C. A linear regression t-test for slope

Step-by-step explanation:

The linear regression t-test for slope is a test for determining the existence of a statistically significant linear relationship between two variables, a independent variable, X and an dependent variable Y in the following form;

Y = B₀ + B₁·X

The test is used to test the regression line slope where we have;

B₀ = The constant

B₁ = The slope also known as the regression coefficient

The given information tested in the question is if an increase in the peaches in the ice cream is associated with an increase in the density of the ice cream, therefore, the test to investigate the relationship between the rate of both variables is the linear regression t-test for slope.

Therefore, we have

Y = B₀ + B₁·X

Where;

X = The percentage of peaches in the ice cream

Y = The density of the ice cream

The error involved in approximating the sum of the given series by the sum of the first five terms of the series is `R5 = 1/6³ + 1/7³ + ...`.

To estimate the error that is made by approximating the sum of the given series by the series the first 5 terms 1/k³, we can use the remainder term of a convergent series.

The given series is ∑ 1/k³ from k = 1 to infinity.We have to find the error involved in approximating the sum of the given series by the sum of the first five terms of the series.

That is, we need to find the difference between the actual sum of the series and the sum of the first five terms of the series.

The sum of the series is given by: `S = 1/1³ + 1/2³ + 1/3³ + 1/4³ + ... + 1/n³ + ...` We can use the remainder term of the series to find the error in approximation.

The remainder term `Rn` is given by: `Rn = Sn - S` where `Sn` is the sum of the first `n` terms of the series. Thus, we have to find the remainder term for `n = 5`.

The remainder term `Rn` is given by: `Rn = S - Sn = 1/6³ + 1/7³ + ...` Since the given series is convergent, the remainder term `Rn` tends to zero as `n` tends to infinity.

So, if we take the sum of the first five terms of the series, the error involved in approximation is given by the remainder term `R5`.

The error involved in this approximation is very small and can be neglected.

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Answer: 15965 kg

Step-by-step explanation:

\(4523+3249+20+1814+2425+3934=15965\)

The area of the half-cylinder can be found using a parametric description of the surface.

To find the area of the half-cylinder, we can parametrize the surface using cylindrical coordinates. Let's consider the surface as a function of two parameters, θ and z. We can define the parametric equations as follows:

x = r cos(θ)

y = r sin(θ)

z = z

In this case, the radius r is given as 4, the angle θ varies from 0 to 2π, and the height z varies from 0 to 7.

To calculate the surface area, we use the formula for the surface area of a surface described by parametric equations:

A = ∫∫ ||rθ × rz|| dθ dz

Here, ||rθ × rz|| represents the magnitude of the cross product of the partial derivatives of the parametric equations with respect to θ and z.

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Answer:

Cause it is more than 50 pounds

Step-by-step explanation:

Answer:

$282

Step-by-step explanation:

There is 60 total number of marbles in the bag for the probability of selecting a red marble is 7/10.

The probability of an event occurring is the fraction of the number of required outcome divided by the total number of possible outcomes.

let the total possible outcome = x

probability of selecting a red marble = P(R) = 7/10

Given that there are 42 red marbles tgen:

42/x = 7/10

x = (42 × 10)/7 {cross multiplication}

x = 420/7

x = 60

Therefore, given the probability of selecting a red marble to be 7/10, the total number of marbles in the bag is derived to be 60

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The value of tan(t−π) is 4/9.

According to the statement

we have given that  tan(t)=4/9 and we have to find the value of tan(t−π).

So,

tan(t−π)  -(1)

take negative sign common from equation (1) it then

tan(t−π) = -tan(-t+π)

and we know that the according to the mathematics formula it become

tan(-t+π) is -tan t

then

tan(t−π) = -(-tan t)

it becomes

tan(t−π) = tan t

then its value becomes

tan(t−π) = tan(t)=4/9.

because we have given that the value of tan t is tan(t)=4/9.

So, The value of tan(t−π) is 4/9.

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Using Definition 1 of the Laplace transform, we have determined the Laplace transforms of the given functions as mentioned above.

Definition 1 of the Laplace transform states that for a function f(t) defined for t ≥ 0, its Laplace transform F(s) is given by F(s) = L{f(t)} = ∫[0,∞] e^(-st) f(t) dt. Using this definition, we can determine the Laplace transforms of the given functions:

1. The Laplace transform of t is given by L{t} = 1/s².

2. The Laplace transform of t² is given by L{t²} = 2/s³.

3. The Laplace transform of e^(6t) is given by L{e^(6t)} = 1/(s - 6).

4. The Laplace transform of te^(3t) requires applying the property of the Laplace transform for the derivative of a function. The Laplace transform of te^(3t) is given by L{te^(3t)} = -d/ds (1/(s - 3)²).

5. The Laplace transform of cos(2t) requires using the trigonometric property of the Laplace transform. The Laplace transform of cos(2t) is given by L{cos(2t)} = s/(s² + 4).

In conclusion, using Definition 1 of the Laplace transform, we have determined the Laplace transforms of the given functions as mentioned above.

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If sample has mean-mass of 2.47 grams and standard-deviation of 0.04 gram, then test-statistic for population mean has (d) a "t-distribution" with 9 degrees-of-freedom.

A "test-statistic" is defined as a value calculated from a sample of data which is used in hypothesis testing.

The test statistic(t) for population mean can be calculated using formula:

⇒ t = (x' - μ)/(s/√n),

where x' = sample mean, μ = population mean, s = sample standard deviation, and n = sample size,

Substituting the values,

We get,

⇒ t = (2.47 - 2.5)/(0.04/√10) ≈ -2.38,

Since the population standard deviation is unknown, we need to use the t-distribution to find the p-value and make a conclusion about the null hypothesis.

The degrees-of-freedom for the t-distribution is = n - 1 = 10 - 1 = 9.

Therefore, the test statistic for the population mean has a t-distribution with 9 degrees of freedom, the correct option is (d).

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The given question is incomplete, the complete question is

The distribution of mass for United States pennies minted since 1982 is approximately normal with mean 2.5 grams. A random sample of 10 pennies minted since 1982 was selected. The sample had a mean mass of 2.47 grams and a standard deviation of 0.04 gram.

The test statistic for the population mean has which of the following distributions?

(a) A normal distribution with mean 0 and standard deviation 1,

(b) A normal distribution with mean 2.5 and standard deviation 0.04,

(c) A normal distribution with mean 2.47 and standard deviation 0.04,

(d) A t-distribution with 9 degrees of freedom,

(e) A t-distribution with 10 degrees of freedom.

If x is a continuous random variable then P(x≥A) =  P(X>A). So, the correct option is C, i.e., P(X>A).

For a continuous random variable, the probability of X taking any specific value is zero, since there are infinitely many possible values that X could take. Instead, we calculate the probability that X falls within a certain range of values.

To find the probability that X is greater than or equal to A, we can use the cumulative distribution function (CDF) of X, denoted as F(X). The CDF F(X) gives the probability that X is less than or equal to a certain value x, i.e., F(X) = P(X ≤ x).

Now, since the event X ≥ A is the complement of the event X < A, we can use the complement rule to write:

P(X ≥ A) = 1 - P(X < A)

Using the definition of CDF, we get:

P(X ≥ A) = 1 - F(A)

Therefore, the probability of a continuous random variable X being greater than or equal to a certain value A is P(X>A) = 1 - F(A). This is the correct option C in the given question.

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Complete question is:

If x is a continuous random variable then P(x≥A)=

A. P(X<A)  

B. 1-P(X>A)

C. P(X>A)

D. 1-P(X≥A-1).

The minimal approximation error A-A 2achievable by a rank-1 approximation A to A is

||A - A1||₂=0 where A is 2×2 matrix.

We have given a matrix A as seen in above figure or A = [ 5 15 ; 6 18 ; -1 -3 ; -4 -12 ; 2 6]

note here that C₂= 3C₁

where Cᵢ --> iᵗʰ column

v = [ 5 ; 6 ; -1 ; -4 ; 2]

||v||² = 82 => ||v|| = √82

and A At v = 820 v

and A At = [ 82 246 ; 246 738]

AAt [1;3] = 820[1;3]

=> v1 = 1/√10(1,3)^t

Then the best rank of 1 approx

= √820 /√80√10 [ 5 ; 6 ; -1 ; -4 ; 2] [ 1 3]

= A

Since , the rank of matrix A is one so, the minimal approx. value is ||A - A1||2 = 0

Hence, the minimal Approx. ||A - A1||2 = 0 .

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Complete question:

Consider the matrix A: A = [5 15] 6 18 -1 -3 -4 -12 [26] What is the minimal approximation error A-A 2 achievable by a rank-1 approximation A to A? Hint: Can you determine this without explicitly calculating the SVD?

Step-by-step explanation:

\(56 = 2n + 99\)

Collect like terms and simplify

\(56 - 99 = 2n \\ - 43 = 2n\)

Divide both sides of the equation by 2

\( \frac{ - 43}{2} = \frac{2n}{2} \)

Simplify

\(n = - \frac{43}{2} = - 21 \frac{1}{2} \)

The value of x in the diagram is 100 degrees

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

The figure

On the figure, we can see that:

The angle x and 100 degrees are alternate angles

Alternate angles have equal values

So, we have

x = 100

Hence, the value is 100 degree

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The 106° measure of arc \(m\widehat{HJ}\) and the length of the chords \(\widehat{FH}\) and \(\widehat{JH}\), which are the same, indicates that m∠G = 21° and ΔFGH is not isosceles. The correct option is option C;

C. The measure of ∠G is 21°, and triangle FGH is not isosceles

An isosceles triangle is a triangle that have two sides of the same length and two angles of the same measure.

According to the outside angle to a circle theorem, the measure of the outside angle ∠G formed by the secant GJ and the tangent GF is equal to the difference of the measures of the arcs \(m\widehat{F'KJ}\), and arc \(m\widehat{FH}\) divided by 2.

\(m\widehat{F'KJ}\) = 360° - \(m\widehat{FH}\) - \(m\widehat{HJ}\)

\(\widehat{FH}\) = \(\widehat{JH}\)

\(\widehat{FH}\) ≅ \(\widehat{JH}\)

The lengths of the chord intercepted by congruent arcs are congruent.

Therefore; \(\widehat{JH}\) ≅  \(\widehat{FH}\)

Which indicates;

\(m\widehat{JH}\) = 106° = \(m\widehat{FH}\)

\(m\widehat{F'KJ}\) = 360° - 106° - 106° = 148°

m∠G = (\(m\widehat{F'KJ}\) - \(m\widehat{FH}\)) ÷ 2

m∠G = (148° - 106°) ÷ 2 = 21°

\(m\widehat{F'KJ}\) = 148°

m∠FHJ = 148° ÷ 2 = 74° (Angle at the center is twice angle formed at the circumference)

∠FHJ = ∠HFG + ∠G (exterior angle to triangle ΔFGH)

∠HFG = ∠FHJ - ∠G

Therefore; m∠HFG = 74° - 21° = 53°

m∠HFG = 53°

m∠FHG = 180° - 74° = 106° (linear pair angles property)

The correct option is option C.

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Answer: A

Step-by-step explanation: hope this helps!

please give brainliest!

Answer:

A. Michael must get a degree in engineering LOLLLLL