The area of a newspaper page​ (opened up) is about 640. 98 square inches. Determine the length and width of the page if its length is about 1. 23 times its width

i think its pi because im smarts thats why

\( \huge \mathbf{Answer࿐}\)

Area of Rhombus :

\(\boxed{ \frac{1}{2} \times diagonal_1 \times diagonal_2}\)

Correct option B. 42 cm²

_____________________________

\(\mathrm{ \#TeeNForeveR}\)

4(6x³ -2x² + 1) is the  polynomial expression  with GCF 4.

Polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and positive-integer powers of variables

The polynomial expression is 24x³ - 8x² + 4

We have to write this polynomial expression in the factored from by taking GCF.

Rewrite 24 as 4×6

8 as 4×2

24x³ - 8x² + 4

4(6x³ -2x² + 1)

Hence, 4(6x³ -2x² + 1) is the  polynomial expression  with GCF 4.

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The percent of attendance figures that differs from the mean by 1500 persons or more is 6.68%.

From the question above, Mean μ = 10,000

Standard Deviation σ = 1,000

The formula for z-score is :

z = (x-μ) / σ

Where, x = observation

z = z-score

Mean μ = 10,000

Standard Deviation σ = 1,000

From the above formula, let's calculate z-score for x = 11,500

z = (x-μ) / σ

z = (11,500 - 10,000) / 1000

z = 1.5

Now, find the probability of attendance figures that differs from the mean by 1500 persons or more.

P(z ≥ 1.5) = 0.0668

To find the percentage, we need to multiply the above value by 100.

P(z ≥ 1.5) × 100 = 0.0668 × 100 = 6.68%

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Answer: Hii <3

11/10 is your answer i believe

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

hopefully this helps you !

- Matthew ~

Answer:

The value of x = 11/10 or 1.1

Step-by-step explanation:

5x+3/-2 =4

=> 5x -3/2 = 4

=> 5x = 4 + 3/2

=> 5x = (8+3)/2

=> 5x = 11/2

=> x = 11/10 or 1.1

The absolute error magnification factor is c) 0.52.

To find the absolute error magnification factor, we need to compare the absolute error in the root approximation with the magnitude of the actual root. The absolute error is given by |\(x_r\) - x|, where \(x_r\) is the root approximation and x is the actual root.

In this case, the root approximation is \(x_r\) = 3.52, and the actual root can be found by solving the equation f(x) = 0:

f(x) = x² - 12x + 27 = 0

Using the quadratic formula, we can solve for x:

x = (12 ± √(12² - 4(1)(27))) / (2(1))

x = (12 ± √(144 - 108)) / 2

x = (12 ± √36) / 2

x = (12 ± 6) / 2

x = 9 or x = 3

Since x = 9 is the actual root, we can calculate the absolute error:

|3.52 - 9| = 5.48

The absolute error magnification factor is then given by |(\(x_r\) - x) / x|:

|(3.52 - 9) / 9| ≈ 0.5133

Rounded to two decimal places, the absolute error magnification factor is approximately 0.52.

Therefore, the correct answer is c. 0.52.

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         Tip: You seem like you are slightly stuck. That is okay! Don't forget the formulas that can help you solve for the circumference. When you can solve for the circumference, the rest of the answers come a bit easier.

[a] 88 inches

         The circumference can be found with C = 2πr, but they tell us to use \(\frac{22}{7}\) instead of pi, so we will do just that.

C = 2 * \(\frac{22}{7}\) * r

C = 2 * \(\frac{22}{7}\) * (14)

C = 88

[b] 176 inches

         Again, the circumference can be found with C = 2πr, but they tell us to use \(\frac{22}{7}\) instead of pi, so we will do just that.

C = 2 * \(\frac{22}{7}\) * r

C = 2 * \(\frac{22}{7}\) * (28)

C = 176

[c] When you double the radius, the circumference also doubles.

         We will divide the second circumference by the first circumference  so we can see how they compare:

176 / 88 = 2

         Now we will take the second radius divided by the first radius:

28 / 14 = 2

         We could use a longer explanation, but in summary, when the radius is doubled the circumference is also doubled as seen with the value 2.

Equation =》s = x² - 3

To change the subject to 'x' :-

\(\tt\:s = x ^ { 2 } - 3\)

\(\tt\:x^{2}-3=s\)

\(\tt\:x^{2}=s+3\)

So,

\(\boxed{\tt\:x=\sqrt{s+3}}\)

\(\boxed{\tt\:x=-\sqrt{s+3}\text{, }s\geq -3}\)

=》'x' can be any of the 2 values in the boxes.

_______

Hope it helps ⚜

Answer:

I agree with RainbowSalt2222

Step-by-step explanation:

It is correct exactly what I got

Answer:

This system has infinite number of solutions.

Step-by-step explanation:

It has 2 variables.

Answer:

35cm3

Step-by-step explanation:

7x5=35

Answer: The correct answer is 35. 7 * 5 = 35

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

PEMDAS

FILLER FILLER FILLER

Answer:

15

Step-by-step explanation:

2x2x2=8

8x2=16

16/16=1

8+8=16

16-1=15

Answer:

110

Step-by-step explanation:

just multiply it by 2

Answer:

110

Step-by-step explanation:

if she paints 55 in half an hour then in a hour she will paint 110 because 55+55=110

The volume of the pot is = 731.2 cubic inches.

Each thing in three dimensions takes up some space. The volume of this area is what is being measured.

The space occupied within an object's borders in three dimensions is referred to as its volume.

It is sometimes referred to as the object's capacity.

Finding an object's volume can help us calculate the quantity needed to fill it, such as the volume of water needed to fill a bottle, aquarium, or water tank.

Since the forms of various three-dimensional objects vary, so do their volumes.

According to our answer-

volume of a pot is = πr²h + 2πr³

22/7 * 698/3

731.2 cubic inches

Hence, The volume of the pot is = 731.2 cubic inches.

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The confidence interval does not contain the hypothesized value of 8 minutes, indicating that the population mean waiting time is likely to be different from 8 minutes. This supports the conclusion that was drawn in part (c), where we failed to reject the null hypothesis

(a) The hypotheses for this application are:

H0: The population mean time a shopper stands in a supermarket checkout line is less than or equal to 8 minutes.

Ha: The population mean time a shopper stands in a supermarket checkout line is greater than 8 minutes.

(b) Given that the sample size (n) is 105, the sample mean waiting time \((\bar X)\) is 8.4 minutes, and the population standard deviation\((\sigma)\) is 3.2 minutes, we can calculate the test statistic and p-value.

The test statistic is calculated using the formula:

\(t = (\bar X - \mu )/ (\sigma / \sqrt n)\)

Plugging in the values, we get:

\(t = (8.4 - 8) / (3.2 / \sqrt {105} ) \approx 1.118\)

To find the p-value, we compare the test statistic to the t-distribution with \((n-1)\)degrees of freedom. In this case, we have \((105-1) = 104\) degrees of freedom. By looking up the p-value associated with the test statistic in the t-distribution table or using statistical software, we find the p-value to be approximately 0.1331.

(c) At \(\alpha = 0.05\), comparing the p-value (0.1331) to the significance level, we find that the p-value is greater than α. Therefore, we do not reject the null hypothesis (H0). There is insufficient evidence to conclude that the population mean waiting time differs from 8 minutes.

(d) To compute a 95% confidence interval for the population mean, we use the formula:

\(CI = \bar X \pm (t_\alpha/2) \times (\sigma / \sqrt n)\)

Plugging in the values, we get:

\(CI = 8.4\pm (1.984 \times (3.2 / \sqrt {105}))\)

\(CI \approx 8.4 \pm0.6438\)

\(CI \approx (7.756, 8.944)\)

The confidence interval does not contain the hypothesized value of 8 minutes, indicating that the population mean waiting time is likely to be different from 8 minutes. This supports the conclusion that was drawn in part (c), where we failed to reject the null hypothesis. The confidence interval provides a range of plausible values for the population mean, and since it does not include 8, it suggests that the mean waiting time is higher than 8 minutes.

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Trevor and Kim will be situated 21 kilometers apart from each other at 1:30 PM. They will be separated by a distance of 21 km when the clock strikes 1:30 in the afternoon.

To determine at what time Trevor and Kim will be 21 km apart, we can set up a distance-time equation based on their relative speeds and distances.

Let's assume that t represents the time elapsed in hours since noon. At time t, Trevor would have traveled a distance of 8t km, while Kim would have traveled a distance of 6t km in the opposite direction.

Since they are running in opposite directions, the total distance between them is the sum of the distances they have traveled:

Total distance = 8t + 6t

We want to find the time when this total distance equals 21 km:

8t + 6t = 21

Combining like terms, we have:

14t = 21

To solve for t, we divide both sides of the equation by 14:

t = 21 / 14

Simplifying, we find:

t = 3 / 2

So, they will be 21 km apart after 3/2 hours, which is equivalent to 1 hour and 30 minutes.

Therefore, Trevor and Kim will be 21 km apart at 1:30 PM.

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Answer:

12x -7

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

(125) ^ 1/3

What number multiplied by itself 3 times is 125

5*5*5

25*5

125

Answer:

5

Step-by-step explanation:

5^3=125

So the cube root of 125 is 5.

Comment: RMS value is equal to the average value. This means that the signal does not have any high-frequency content. It can be inferred that the function y(t) does not oscillate. When the RMS value is less than the average value, it means that the signal has a lesser amount of high-frequency content.

Average and rms values for the function y(t)=25+10sin6πt over the time periods (a) 0 to 0.1 s, (b) 0.4 to 0.5 s, (c) 0 to 1/3 s, and (d) 0 to 20 s are as follows:

a) For t=0 to t=0.1s:

Average value, y_avg = 25

RMS value, y_RMS = 25.1987

Comment: RMS value is greater than the average value. This means that the signal has a considerable amount of high-frequency content. It can be inferred that the function y(t) oscillates rapidly.

b) For t=0.4 to t=0.5s:

Average value, y_avg = 25

RMS value, y_RMS = 28.2843

Comment: RMS value is greater than the average value. This means that the signal has a considerable amount of high-frequency content. It can be inferred that the function y(t) oscillates rapidly.

c) For t=0 to t=1/3 s:

Average value, y_avg = 25

RMS value, y_RMS = 23.7176

Comment: RMS value is less than the average value. This means that the signal has a lesser amount of high-frequency content. It can be inferred that the function y(t) oscillates slowly.

d) For t=0 to t=20 s:

Average value, y_avg = 25

RMS value, y_RMS = 25

Comment: RMS value is equal to the average value. This means that the signal does not have any high-frequency content. It can be inferred that the function y(t) does not oscillate. Comment on the nature and meaning of the results in terms of analysis of dynamic signals.The results indicate that the function y(t) oscillates rapidly at the start and end of the time period and slowly in the middle. When the RMS value is greater than the average value, it means that the signal has a considerable amount of high-frequency content.

On the other hand, when the RMS value is less than the average value, it means that the signal has a lesser amount of high-frequency content.

Furthermore, if the RMS value is equal to the average value, it means that the signal does not have any high-frequency content.

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Answer:

Yes

Step-by-step explanation:

An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.

Answer:

yes

Step-by-step explanation:

Answers:

\((f \circ g)(x) = 2x^2-5x\\\\(g \circ f)(x) = 2x^2+7x\\\\\)

The domain is "all real numbers" which you could type in (-infinity, infinity) or \((-\infty, \infty)\) when doing interval notation. This applies to both.

=========================================================

Explanation:

See the attached image for the steps of each.

The domain is the set of all real numbers because each result is a polynomial. We don't have to worry about dividing by zero, taking a square root of a negative number etc. There are no restrictions on x. Any real number can replace x to get some real number result for y. This applies to both composite functions.

The probability of getting at least one green card is 55/64.

Part (a)A tree diagram can help to keep track of the possibilities when drawing two cards with replacement from a deck of eight cards.

In this case, we have two events: G1 = first card is green G2 = second card is green The tree diagram for the given problem is as shown below: 5/8 G 3/8 Y 5/8 G 3/8 Y 5/8 G 3/8 Y G1 G1 Y G1 G2 G2 G2 G2

Part (b) Probability of first card being green P(G1) = 5/8 Probability of second card being green given that the first card was green P(G2|G1) = 5/8

So, P(G1 and G2) = P(G1) x P(G2|G1) = 5/8 x 5/8 = 25/64

Therefore, P(G1 and G2) = 25/64

Part (c)Probability of getting at least one green card means the probability of getting one green card and the probability of getting two green cards.

P(at least one green) = P(G1 and Y2) + P(Y1 and G2) + P(G1 and G2) P(at least one green)

= P(G1) x P(Y2) + P(Y1) x P(G2) + P(G1) x P(G2|G1) P(at least one green)

= (5/8) x (3/8) + (3/8) x (5/8) + (5/8) x (5/8)

= 15/64 + 15/64 + 25/64

= 55/64

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For a Poisson process with rate λ, the interarrival times between events are exponentially distributed with parameter μ = 1/λ. So, the time between the (n-1)-th and n-th event, denoted as Tn, follows an exponential distribution with parameter μ = 1/3 minutes.

Since Sn is the sum of the first n interarrival times, we have:

Sn = T1 + T2 + ... + Tn

The sum of n exponential random variables with parameter μ is a gamma random variable with shape parameter n and scale parameter μ. Therefore, Sn follows a gamma distribution with shape parameter n and scale parameter μ.

In this case, n = 10 and μ = 1/3. So, E[S10] can be calculated as:

E[S10] = n * μ = 10 * (1/3)

= 10/3 minutes.

Therefore, E[S10] = 10/3 minutes.

b. E[S4|N(1) = 3]:

Given that N(1) = 3, we know that there are 3 events in the first minute. Therefore, the time of the 4th event, S4, will be the sum of the first 3 interarrival times plus the time between the 3rd and 4th event.

Using the same reasoning as in part a, we know that the sum of the first 3 interarrival times follows a gamma distribution with shape parameter 3 and scale parameter 1/3. The time between the 3rd and 4th event, denoted as T4, follows an exponential distribution with parameter 1/3.

So, S4 = T1 + T2 + T3 + T4.

Since T1, T2, T3 are independent of T4, we can calculate E[S4|N(1) = 3] as:

E[S4|N(1) = 3] = E[T1 + T2 + T3 + T4]

= E[T1 + T2 + T3] + E[T4]

= (3/3) + (1/3)

= 4/3 minutes.

Therefore, E[S4|N(1) = 3] = 4/3 minutes.

c. Var(S10):

The variance of Sn, Var(Sn), for a Poisson process with rate λ, is given by:

Var(Sn) = n * σ^2,

where σ^2 is the variance of the interarrival times.

In this case, n = 10 and the interarrival times are exponentially distributed with parameter μ = 1/3. The variance of an exponential distribution is \(\mu^2\)So, \(\sigma^2 = \left(\frac{1}{3}\right)^2\)

= 1/9.

Substituting the values into the formula, we have:

Var(S10) = 10 * (1/9)

= 10/9.

Therefore, Var(S10) = 10/9.

d. E[N(4) − N(2)|N(1) = 3]:

Given that N(1) = 3, we know that there are 3 events in the first minute. Therefore, at time t = 2 minutes, there will be 3 - 1 = 2 events that have already occurred.

Now, we need to find the expected value of the difference in the number of events between time t = 4 minutes and t = 2 minutes, given that there were 3 events at t = 1 minute.

Since the number of events in a Poisson process follows a Poisson distribution with rate λt, where t is

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a) The characteristics of the two events "attends their Bus-230 weekly meeting" and "does not attend their Bus-230 weekly meeting" are as follows:

1. Mutually Exclusive: The two events are mutually exclusive, meaning that an individual can either attend the Bus-230 weekly meeting or not attend it. It is not possible for someone to both attend and not attend the meeting at the same time.

2. Collectively Exhaustive: The two events are collectively exhaustive, meaning that they cover all possible outcomes. Every individual either attends the meeting or does not attend it, leaving no other possibilities.

b) Based on the characteristics described in part a), we can conclude the following about the probability of these two events:

1. The sum of the probabilities: Since the two events are mutually exclusive and collectively exhaustive, the sum of their probabilities is equal to 1. In other words, the probability of attending the meeting (Pl'attends their Bus-230 weekly meeting) plus the probability of not attending the meeting (Pl' does not attend their Bus-230 weekly meeting) equals 1.

2. Complementary Events: The two events are complementary to each other. If we know the probability of one event, we can determine the probability of the other event by subtracting it from 1. For example, if the probability of attending the meeting is 0.7, then the probability of not attending the meeting is 1 - 0.7 = 0.3.

These conclusions are based on the fundamental properties of probability and the characteristics of mutually exclusive and collectively exhaustive events.

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Answer:

32%

Step-by-step explanation:

Area of rectangle ABCD = 20 * 30 = 600 Sq cm

After increasing the length and width of the rectangle by 20% and 10% respectively.

New length = 30 + 20% of 30 = 30 + 6 = 36 cm

New width = 20 + 10% of 20 = 20 + 2 = 22 cm

Area of new rectangle = 36*22 = 792 Sq cm

Increase in area = 792 - 600 = 192 Sq cm

Percentage increase in area

= (192/600) *100

= 32%

Answer:

nnnnnnnfjnjzfv

Step-by-step explanation:

The number of combination 6-letter words using five vowels in alphabetical order,

5*4*3*2*1*21=2520

Permutation and combination are methods for representing a collection of objects by selecting them from a set and forming subsets. It specifies the various ways to arrange a specific set of data. Permutations are when we select data or objects from a specific group, whereas combinations are the order in which they are represented.

The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations. Combination refers to the combination of n things taken k at a time without repetition. To refer to combinations in which repetition is allowed, the terms k-selection or k-combination with repetition are often used.

We have five vowels: a, i, e,0, and u.

make the 6-letter word using five vowels in alphabetical order,

5*4*3*2*1*21=2520

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The mass of the solid with density p(x,y,z) = 16, bounded by z = x² + y² and z = 1 is 2 units

To find the mass of the solid, we need to integrate the density function over the volume of the solid. In this case, the solid is bounded by the surfaces z = x² + y² and z = 1.

So, we need to set up a triple integral in cylindrical coordinates. The bounds of integration are

0 ≤ r ≤ 1 (since the solid is bounded by z = 1)

0 ≤ θ ≤ 2π (the solid is symmetric around the z-axis)

r² ≤ z ≤ 1 (since the solid is bounded by z = x² + y²)

Therefore, the mass of the solid is given by:

M = ∫∫∫ p(x,y,z) dV

= ∫∫∫ 16 r dz r dr dθ

= 16 ∫∫ (1 - r²) r dr dθ (limits: 0 to 1 for r and 0 to 2π for θ)

= 16 ∫0^1 ∫0^2π (r - r³) dθ dr

= 16 ∫0^1 (r - r³) dr

= 16 [1/2 r² - 1/4 r⁴] from 0 to 1

= 16/8

= 2 units

Therefore, the mass of the solid is 2 units

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9514 1404 393

Answer:

  see attached for a plot

  14.44, 15.21, 3.8727² ≈ 14.99780529

Step-by-step explanation:

(3.8)² = 14.44

(3.9)² = 15.21

The root of 15 will be approximately at the spot where the line between these values crosses 15.

We can see the slope of the square function is approximately ...

  m = (15.21 -14.44)/(3.9 -3.8) = 0.77/0.1 = 7.7

Then the value we need to add to 3.8 will be ...

  (15.00 -14.44)/7.7 = 0.56/7.7 ≈ 0.0727

An approximation of the root is 3.8727.

_____

Additional comment

Here, the location we plotted for √15 is "exact." We're not sure what original approximation you're trying to better. We chose to use linear interpolation between the points (3.8, 14.44), (3.9, 15.21) to estimate the value of 'x' that would give x^2 = 15. (There are other, better, ways to refine the estimate of the root.)

Answer:

30 sq units

3 + 2 = 5

5 x 6 = 30

As, stx = tsx for every eigenvector x of t, and eigenvectors corresponding to distinct eigenvalues are linearly independent, we can conclude that st = ts.

To prove that st = ts, where v is finite-dimensional, t and s are linear operators on v, t has dim v distinct eigenvalues, and s has the same eigenvectors as t (not necessarily with the same eigenvalues), we can use the fact that eigenvectors corresponding to distinct eigenvalues are linearly independent.

Let's consider an eigenvector x of t with eigenvalue λ. We can write this as tx = λx. Now, since s has the same eigenvectors as t, we can write this as sx = λx.

Now, let's consider the product stx. Using the definitions of s and t, we have stx = s(λx) = λ(sx).

Since sx = λx, we can substitute this in the above equation to get stx = λ(λx) = λ²x.

On the other hand, let's consider the product tsx. Using the definitions of s and t, we have tsx = t(λx) = λ(tx).

Since tx = λx, we can substitute this in the above equation to get tsx = λ(λx) = λ²x.

Since stx = tsx for every eigenvector x of t, and eigenvectors corresponding to distinct eigenvalues are linearly independent, we can conclude that st = ts.

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