Compare: 0.06 x 10-1 O 10-2A>B)

Answer :

a) It is given that

Everyday a machine makes 500000 staples and puts them into the boxes.

The machine need 170 staples to fill a box

One box contans = 170 staples.

Total number of staples = 500000

Number of box required to fill 500000 staples = Total number of staples/No. of staples 1 box contains.

\( \: :\implies \) 500000 /170

\( :\implies \: \) 2941 (approx)

Therefore, 2941 boxes are required to fill with 500000 staples.

b) It is given that,

Each staple is made of 0.21 g of metal .

Total weight of metal = 1 kg = 1000 g

1 staple = 0.21 g

Number of staples = weight of metal/ Weight of 1 staple.

\( \: :\implies \) 1000/0.21

\( \: :\implies \) 4761 (approximately)

Therefore, 4761 staples can be made from 1 kg of the metal.

Answer:

Step-by-step explanation:

tan30 = opp/adj = x/1

x = tan30 = ⅓√3 or approximately 0.577

a ) Construction of stem and leaf display of the data is:

For n = 129 and with splint unit = 0.1, the stem and splint map of the given data on Shower- inflow rate( L/ min) is as follows

a stem-and-leaf display of the data.

   Stems                                                Leaves

       2                                                       28

       3                                                1344567789

       4                                                 01356889

       5                                          00001114455666789

       6                              0000122223344456667789999

       7                                     00012233455555668

       8                                                02233448

       9                                         012233335666788

      10                                             2344455688

      11                                                2335999

      12                                                     17

      13                                                      9

      14                                                     36

      15                                                  0035

      16                                                  None

      17                                                  None

      18                                                     3

b) From brume and splint map we note that minimal Shower inflow rate is2.2 whereas outside is18.3 L/ mim. farther typical or representative rate is7.0 L/min.

c) The display of data on steam and leaf chart shows that data is positively skewed means concentration of data on left side or lower value side is high as compared to other side.

d) Distribution is not symmetric rather very clear positive skew ness is observed through steam and leaf chart. Even distribution is Unimodal.

e) From steam and leaf chart is indicative to conclude that the highest observation 18.3 is outlier.

A stem and splint plot, also known as a stem and splint illustration, is a way to arrange data so that it's simple to see how constantly colorful feathers of values do. It's a graph that displays ordered numerical data. A stem and a splint are divided into each piece of data.

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Complete question:

The accompanying data set consists of observations on shower-flow rate (L/min) for a sample of n = 129 houses:

4.6 12.1 7.1 7.0 4.0 9.2 6.7 6.9 11.5 5.1

11.2 10.5 14.3 8.0 8.8 6.4 5.1 5.6 9.6 7.5

7.5 6.2 5.8 2.8 3.4 10.4 9.8 6.6 3.7 6.4

8.3 6.5 7.6 9.3 9.2 7.3 5.0 6.3 13.9 6.2

5.4 4.8 7.5 6.0 6.9 10.8 7.5 6.6 5.0 3.3

7.6 3.9 11.9 2.2 15.0 7.2 6.1 15.3 18.3 7.2

5.4 5.5 4.3 9.0 12.7 11.3 7.4 5.0 3.5 8.2

8.4 7.3 10.3 11.9 6.0 5.6 9.5 9.3 10.4 9.7

5.1 6.7 10.2 6.2 8.4 7.0 4.8 5.6 10.5 14.6

10.8 15.5 7.5 6.4 3.4 5.5 6.6 5.9 15.0 9.6

7.8 7.0 6.9 4.1 3.6 11.9 3.7 5.7 6.8 11.3

9.3 9.6 10.4 9.3 6.9 9.8 9.1 10.6 4.5 6.2

8.3 3.1 4.9 5.0 6.0 8.2 6.3 3.8 6.0

(a) Construct a stem-and-leaf display of the data. (Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)

Stems Leaves

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

(b) What is a typical, or representative, flow rate?

L/min

(c) Does the display appear to be highly concentrated or spread out?

highly concentrated, except for a few values on the positive sidehighly concentrated in the middle highly concentrated, except for a few values on the negative sidespread out

(d) Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry?

Yes, the distribution appears to be reasonably symmetric.No, the data are skewed to the right, or positively skewed. No, the data are skewed to the left, or negatively skewed.No, the distribution of the values appears to be bimodal.

(e) Would you describe any observation as being far from the rest of the data (an outlier)?

Yes, the value 2.2 appears to be an outlier.Yes, the value 15.5 appears to be an outlier. Yes, the value 18.3 appears to be an outlier.No, none of the observations appear to be an outlier.

Velocity is not constant over time and can be influenced by a variety of factors. Changes in velocity can have implications for the relationship between changes in the money supply and the price level, highlighting the complexity of monetary dynamics in an economy.

According to the quantity equation (MV = PQ), changes in the money supply (M) will lead directly to changes in the price level (P) if velocity (V) and real GDP (Q) remain constant. However, velocity is not necessarily constant over time and can be influenced by various factors.

Velocity represents the rate at which money circulates in the economy, indicating how quickly money is used to facilitate transactions. It is influenced by factors such as changes in consumer and business behavior, technological advancements, financial innovation, and shifts in confidence and trust in the economy.

Changes in velocity can occur due to several reasons. For example:

Changes in Transaction Patterns: Shifts in consumer preferences or business practices can alter the frequency and speed of transactions, affecting how money circulates and influencing velocity.

Financial Innovation: Advancements in payment systems, such as the rise of electronic transactions, online banking, or mobile payments, can potentially impact the velocity of money by facilitating faster and more efficient transactions.

Confidence and Economic Stability: Changes in economic conditions, inflation expectations, or financial stability can influence people's willingness to hold money, affecting velocity.

Monetary and Fiscal Policies: Changes in monetary policy, such as interest rate adjustments or changes in the money supply, can indirectly affect velocity by influencing borrowing and spending behavior.

While changes in velocity can occur independently of changes in the money supply, they can also be related. For instance, if the money supply increases without a corresponding increase in the demand for money (velocity decreases), it can put upward pressure on prices. On the other hand, if velocity increases due to changes in transaction patterns or increased economic activity, it can offset the impact of changes in the money supply on prices.

Overall, velocity is not constant over time and can be influenced by a variety of factors. Changes in velocity can have implications for the relationship between changes in the money supply and the price level, highlighting the complexity of monetary dynamics in an economy.

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Considering the definition of parallel line, the equation of line l is y= -5x -11.

A linear equation o line can be expressed in the form y = mx + b

where

Parallel lines are two lines that are always at the same distance from each other.

Two lines are parallel if they have the same slope and different y-intercepts. This is, they have the same value of m and different value of b.

In this case, the line is y−3=−5(x−2) Expressed in the form y = mx + b, you get:

y−3=−5x−2×(-5)

y−3=−5x+ 10

y=−5x+ 10 + 3

y=  -5x +13

where:

Two lines are parallel if they have the same slope. So, the parallel line has a slope of -5 and has a form of: y= -5x + b

The line passes through the point (3, -26). Replacing in the expression for parallel line:

-26= -5× 3 + b

-26= -15 + b

-26 + 15 = b

-11 = b

Finally, the equation of parallel line is y= -5x -11.

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Answer:

  78,322

Step-by-step explanation:

We can write the proportion ...

  total/100% = 22,400/28.6%

  total = 78,322 . . . . carry out the division

The student is correct and for the given linear equation is 5.

A linear equation in one variable is one that is written in the form ax + b = 0, where a, and b are real numbers and the coefficients of x and constant, i.e. a and b, are not equal to zero.

A linear equation is an algebraic equation of the form y=mx+b, involving only a constant and a first-order (linear) term, where m is the slope and b is the y-intercept. The preceding is sometimes referred to as a "linear equation of two variables," where y and x are the variables.

Given:

here is a linear equation in which one variable is given.

n - 3 = \(\frac{16}{8}\)

According to the question, on the simplification of the given equation.

\(n - 3 = \frac{16}{8} \\n - 3 = 2\\n = 2 + 3\\n = 5\)

so, n = 5

Hence, the student is correct.

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n - 3 = \(\frac{16}{8}\) The student calculated n = 5. Is the student correct.?

Answer:

yes, a flag is created by sewing it together

Answer:

Slope= -9

Step-by-step explanation:

To find the slope, you need to find two points on the line, so it would be (-1,4) and (0,-5)

You can use the formula and substitute the coordinates

Y2-Y1/X2-X1

so...

-5-4/0-(-1)

-9/1=-9

or you can find the distance between both points and simplify if needed.

So you move 9 down, and one to the right

Answer:

y = 3/5x - 1

Step-by-step explanation:

y = mx + b

m = slope

m = 3/5

to find the y - intercept we have to make x = 0 by adding or subtracting

in this case it is +5

add five to both x and y

\((-5, -6) + (5, 5) = (0, -1)\)

the y intercept focuses on what y is when x is 0

the y intercept in this case is -1

If the data were transformed and points with the coordinates

(x, log(y)) were plotted then the points (1, 1.079), (2, 1.919), (3, 2.763) were plotted.

To plot the points with the coordinates (x, log(y)), we need to take the logarithm of the y-values.

Let's calculate the logarithms for the given data:

x     y            log(y)

1     12          log(12) = 1.079

2     83        log(83) = 1.919

3     580      log(580) = 2.763

Therefore, the points that would be plotted are (1, 1.079), (2, 1.919), (3, 2.763)

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Answer:

Oil needed for 1 cake = 1/4

So, for 5 cakes = 1/4 * 5

=> 5/4

Therefore you will require 5/4 cups of oil.

If my answer helped, kindly mark me as the brainliest!!

Thank You!!

Answer: Yes, they will have it mowed in 1 hour and 18 minutes

Step-by-step explanation:

The required values of the given situation are k = 0.83 if b varies directly as m, k = 30 if f varies inversely as p, and k = 0.43 if u varies jointly with q and d.

The equation is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are equal.

Since b varies directly as m.

So the equation of variation is b = kx

A study shows that m = 6 when b = 5.

This means 5 = k(6)

So k = 5/6

k = 0.83  (round to two decimal places)

Since f varies inversely as p.

So the equation of variation is f = k/p

It has been observed that p = 6 when ƒ = 5.

So 5 = k/6

k = 30

Since u varies jointly with q and d.

So the equation of variation is u = kqd

It is found that q = 7 and d = 2 when u = 6.

So 6 = k(7)(2) = 14k

k = 6/14 = 3/7 = 0.43 (round to two decimal places)

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72 mph = 105.6 ft/second

In order to convert any value from mph the following formula can be referred to:

speed in miles per hour x  5,280 = speed in foot per second

Therefore when 72 miles per hour is converted to foot per second would equal to the following value:

105.6 feet per second

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The general solution of the partial differential equation is:

U(x, t) = Σ [Aₙ*sin((nπ/3)x)]*e^(-(nπ/3)²t)

The partial differential equation (PDE) is given as:

Uxx = (1/2)Ut with the boundary and initial conditions as 0< X <3 with U(0,t) = U(3, t)=0, U(0, t) = 5sin(4πx)

Assume that the solution can be written as a product of two functions:

U(x, t) = X(x)T(t)

Substituting this into the PDE, we have:

X''(x)T(t) = (1/2)X(x)T'(t)

Dividing both sides by X(x)T(t), we get:

(X''(x))/X(x) = (1/2)(T'(t))/T(t)

Since the left side only depends on x and the right side only depends on t, both sides must be equal to a constant, denoted as -λ²:

(X''(x))/X(x) = -λ²

(1/2)(T'(t))/T(t) = -λ²

Simplifying the second equation, we have:

T'(t)/T(t) = -2λ²

Solving the second equation, we find:

T(t) = Ce^(-2λ²t)

Applying the boundary condition U(0, t) = 0, we have:

U(0, t) = X(0)T(t) = 0

Since T(t) ≠ 0, we must have X(0) = 0.

Applying the boundary condition U(3, t) = 0, we have:

U(3, t) = X(3)T(t) = 0

Again, since T(t) ≠ 0, we must have X(3) = 0.

Therefore, we can conclude that X(x) must satisfy the following boundary value problem:

X''(x)/X(x) = -λ²

X(0) = 0

X(3) = 0

The general solution to this ordinary differential equation is given by:

X(x) = Asin(λx) + Bcos(λx)

Applying the initial condition U(x, 0) = 5*sin(4πx), we have:

U(x, 0) = X(x)T(0) = X(x)C

Comparing this with the given initial condition, we can conclude that T(0) = C = 5.

Therefore, the complete solution for U(x, t) is given by:

U(x, t) = Σ [Aₙsin(λₙx) + Bₙcos(λₙx)]*e^(-2(λₙ)²t)

where:

Σ represents the summation over all values of n

λₙ are the eigenvalues obtained from solving the boundary value problem for X(x).

To find the eigenvalues λₙ, we substitute the boundary conditions into the general solution for X(x):

X(0) = 0: Aₙsin(0) + Bₙcos(0) = 0

X(3) = 0: Aₙsin(3λₙ) + Bₙcos(3λₙ) = 0

From the first equation, we have Bₙ = 0.

From the second equation, we have Aₙ*sin(3λₙ) = 0. Since Aₙ ≠ 0, we must have sin(3λₙ) = 0.

This implies that 3λₙ = nπ, where n is an integer.

Therefore, λₙ = (nπ)/3.

Substituting the eigenvalues into the general solution, we have:

U(x, t) = Σ [Aₙ*sin((nπ/3)x)]*e^(-(nπ/3)²t)

where Aₙ are the coefficients that can be determined from the initial condition.

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The distance from Kano to Lagos is 1200 km

The problem involves addition of fractions

The fraction required are:

let the distance form kano to Lagos be x

it was discovered that a driver has travelled 3/8 = 3x/8

another driver had travelled 3/10 of that distance = 3x/10

The drivers are still 390 km apart

total distance, x

x = 390 + 3x/8 + 3x/10

x - 3x/8 - 3x/10 = 390

0.325x = 390

x = 390 / 0.325

x = 1200 km

From the driving distance information 1200 km is the distance from Kano to Lagos

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Answer:

Answer: Add 315 gallons of 70% antifreeze to 90 gallons of 25% antifreeze to make a 60% mixture.

You need to end up a mixture that is 60% antifreeze.

You have 90 gallons of 25% antifreeze, which means it has .25*90 = 22.5 gallons of pure antifreeze mixed with a solvent (probably water).

.

You will add 'x' gallons of 70% antifreeze to make the 60% mixture.

So, at the conclusion, you will have (90+x) gallons  60% 'pure' antifreeze. That can be shown with the equation:

.

.25*90 + .7*x = .6*(90+x)

.

Multiply by 100 to eliminate fractions.

.

25*90 + 70x = 60(90 +x)

.

2250 + 70x = 5400 + 60x

.

10x = 5400 -2250

.

10x = 3150

.

x = 315 gallons of 70% antifreeze.

.

How much will you end up having on hand?

90 + 315 = 405 gallons

.

If it is 60% antifreeze, how much 'pure' antifreeze will be in the mixture?

.6*405 = 243

.

In the original 90 gallons, you have 22.5 gallons of pure antifreeze.

In the additional 315 gallons, you have .7*315 = 220.5 gallons.

22.5+220.5 = 243 gallons

Step-by-step explanation:

The numbers 3.4, 4.7, 30, 4 1/2, and 6 are solutions to both the inequality and the problem situation. The number -5 is not a solution since time cannot be negative.

The inequality that represents this situation is:

500 ≤ 248 + 62t

Now let's evaluate each number to determine if it is a solution to both the inequality and the problem situation or to the inequality only, or if it is not a solution:

-5: Not a solution (since time cannot be negative)

3.4: Solution to both the inequality and the situation

4.7: Solution to both the inequality and the situation

30: Solution to both the inequality and the situation

4 1/2: Solution to both the inequality and the situation

6: Solution to both the inequality and the situation

Therefore, the numbers 3.4, 4.7, 30, 4 1/2, and 6 are solutions to both the inequality and the problem situation. The number -5 is not a solution since time cannot be negative.

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Answer:

I don't know the answer but I know it is not AB II DC because I did this assignment and got this wrong. This is just to narrow your answer choices down, sorry if this isn't helpful.

Step-by-step explanation:

The line AD is parallel to the line BC. Therefore, option B is the correct answer.

Angles in parallel lines are angles that are created when two parallel lines are intersected by another line called a transversal.

Given that,  m∠3 = m∠B

The corresponding angles definition tells us that when two parallel lines are intersected by a third one, the angles that occupy the same relative position at each intersection are known to be corresponding angles to each other.

Here, m∠3 and m∠B are corresponding angles and they are equal in measure.

So, AD || BC

Therefore, option B is the correct answer.

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Answer:

Im pretty sure its x< 3

Step-by-step explanation:

Answer:

the answer is A 3>x hope this helps

Answer:

(9, 4).

Step-by-step explanation:

To find the corresponding point of the function f(x)+5 when (9,-1) is on the graph of f(x), we simply add 5 to the y-coordinate of the point.

If (9,-1) is on the graph of f(x), then the corresponding point on the graph of f(x)+5 would be (9, -1+5), which simplifies to (9, 4).

Therefore, the corresponding point of the function f(x)+5 is (9, 4).

The best assessment of the linearity of the relationship between the day of the month and account balance would be "Because the residual plot has no obvious pattern, and the scatterplot appears linear, it is appropriate to use the line of best fit to predict the account balance based on the day of the month."The correct answer is option C.

When assessing linearity, it is important to examine both the scatterplot and the residual plot. The scatterplot is used to visualize the relationship between the variables, while the residual plot helps us assess the appropriateness of a linear model by examining the pattern of the residuals (the differences between observed and predicted values).

If the scatterplot appears reasonably linear, it suggests that there is a linear relationship between the variables. In this case, since the scatterplot appears linear, it supports the use of a linear model to predict the account balance based on the day of the month.

Furthermore, the residual plot is used to check for any patterns or systematic deviations from the line of best fit. If the residual plot exhibits no obvious pattern and the residuals appear randomly distributed around zero, it indicates that the linear model captures the relationship adequately.

Therefore, if the residual plot shows no obvious pattern, it provides further evidence in favor of using the line of best fit to predict the account balance based on the day of the month.

In conclusion, when the scatterplot appears linear and the residual plot shows no obvious pattern, it is appropriate to use the line of best fit to predict the account balance based on the day of the month.

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Answer:

person above

Step-by-step explanation:

the obvious

Answer:

If Joseph was going to buy a large number of phones, and he bought the warranty on each phone,  he should expect to lose value from the warranties.

Step-by-step explanation:

Given - Joseph just bought a brand new cell phone and a warranty for the cell phone. The warranty cost 300 euros and  is worth 800 euros if his phone breaks. Joseph estimates that there is a 20% chance of his phone breaking.

Joseph calculates the expected value of buying the warranty to be -140.

To find - Which of the following statements are true?

Joseph will certainly lose value from the warranty.

Joseph will certainly gain value from the warranty.

If Joseph was going to buy a large number of phones, and he bought the warranty on each phone,  he should expect to lose value from the warranties.

Proof -

Given that , expected value = -140

⇒It means Joseph lose 140

(negative represents loss )

Now,

Given the expected values, it means we have given the probability so we can not be sure about loosing money

So,

The 1st and 2nd option is wrong because certain means sure because in the question we are not sure that Joseph phone will break or not.

Now,

The 3rd option says that if Joseph buy large amount of phone then he expect to lose value from from warranties .

This is the correct option because if we buy phones and warranty on phone then the expected value will divide on each phone.

So, the chance of impact on loss will decrease.

The correct option is - If Joseph was going to buy a large number of phones, and he bought the warranty on each phone,  he should expect to lose value from the warranties.

Answer:

look like the answer is 39

Step-by-step explanation:

Answer: your answer is y+b=39-dy 40=4=

80Step-by-step explanation:

Answer:

50.75

Step-by-step explanation:

We have:

\(E[g(x)] = \int\limits^{\infty}_{-\infty} {g(x)f(x)} \, dx \\\\= \int\limits^{1}_{-\infty} {g(x)(0)} \, dx+\int\limits^{6}_{1} {g(x)\frac{2}{x} } \, dx+\int\limits^{\infty}_{6} {g(x)(0)} \, dx\\\\= \int\limits^{6}_{1} {g(x)\frac{2}{x} } \, dx\\\\=\int\limits^{6}_{1} {(4x+3)\frac{2}{x} } \, dx\\\\=\int\limits^{6}_{1} {(4x)\frac{2}{x} } \, dx + \int\limits^{6}_{1} {(3)\frac{2}{x} } \, dx\\\\=\int\limits^{6}_{1} {8} \, dx + \int\limits^{6}_{1} {\frac{6}{x} } \, dx\\\\\)

\(=8\int\limits^{6}_{1} \, dx + 6\int\limits^{6}_{1} {\frac{1}{x} } \, dx\\\\= 8[x]^{^6}_{_1} + 6 [ln(x)]^{^6}_{_1}\\\\= 8[6-1] + 6[ln(6) - ln(1)]\\\\= 8(5) + 6(ln(6))\\\\= 40 + 10.75\\\\= 50.74\)

Answer:

0

Step-by-step explanation:

Step 1: Define

\(\lim_{x \to \infty} e^{x-x^2}\)

Step 2: Substitute in infinity

\(\lim_{x \to \infty} e^{\infty-\infty^2}\)

We know that the second exponent is always going to be bigger than the 1st exponent, so we would get a negative exponent.

\(\lim_{x \to \infty} e^{-\infty}\)

We can simplify a negative exponent into a fraction.

\(\lim_{x \to \infty} \frac{1}{e^{\infty}}\)

We know that if we keep plugging in larger numbers for x, the denominator is going to be infinitely big. And bigger the number we are dividing by, the smaller the entire fraction is going to be.

Therefore, we would be approaching 0.

Answer:

2112

Step-by-step explanation:

100% sure

Answer:

On a number line, −2 is located to the right of −3

Explanation:

It's better constructing a graph

In any case:

A.) On a number line, −2 is located to the left of −3.

-This is wrong. -2 is closer to the zero on the right thus, -3 is on the left of -2 and -2 is on the right of -3

B.) On a number line, −2 is located to the right of −3.

-This is correct

3.) On a number line, −2 is located to the left of 0 and −3 is located to the right of 0.

-This is wrong. Since both -2 and -3 are negative integers, both must be in the left of 0

4.) On a number line, −2 is located to the right of 0 and −3 is located to the left of 0.

-This is wrong. Since both -2 and -3 are negative integers, both must be in the left of 0