6.35x10^7 as a ordinary number

Answer:

                                                 Class

                                    YES                  NO             TOTAL

                    YES         26                      9                   35

Country

                     NO         37                     26                  63

                TOTAL        63                     35                  98

The value of the give expression is  \(\frac{313}{748845}\) OR 4.18 × 10⁻⁴

From the question, we are to determine the value of

2(5+34-67/45*43)^2

This can be written as

\(2 (\frac{5+34-67}{45 \times 43 })^{2}\)

First, we will simplify the bracket

\(2 (\frac{-28}{1935 })^{2}\)

Then, we get

\(2 \times \frac{-28}{1935 }\times \frac{-28}{1935 }\)

= \(\frac{1568}{3744225}\)

= \(\frac{313}{748845}\) OR 4.18 × 10⁻⁴

Hence, the value of the give expression is  \(\frac{313}{748845}\) OR 4.18 × 10⁻⁴

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Answer:

Step-by-step explanation:

a) PQ' / PQ = 12/8 = 3/2

  Q'R' / QR = 15/10 = 3/2

  PR' / PR = 9/6 = 3/2

b) this means that triangle PQR is congruent to triangle PQ'R'. Because if we divide each side we get 3/2 (SSS)

c) because QR and Q'R' are parallel and they have same angles.

now please mark me brainliest as you promised

Answer:

A:33

B:22

C:74

Step-by-step explanation:

A:

2x2x2=8

5x5=25

8+25=33

B:

7x7=49

3x3x3=27

49-27=22

C:

8x8=64

square root of 100 is 10

64+10=74

i hope this helps :)

Answer:

2.5 or 2 1/2

Step-by-step explanation:

i caculated do order of operations

The probability that a student owns a car or a computer is 0.45 and the probability that a student does not own a car or a computer is 0.

So, we know that:

Students who own a car: 0.65 ⇒ 65

Students who own a Computer: 0.82 ⇒ 82

Students who own both: 0.55 ⇒ 55

Probability formula: Favourable events/Total events

The probability that a student owns a car or a computer:

Favourable events: 100 - 55 = 45

P(E) = Favourable events/Total events

P(E) = 45/100

P(E) = 0.45

The probability that a student does not own a car or a computer:

Favourable events: 0

P(E) = Favourable events/Total events

P(E) = 0/100

P(E) = 0

Therefore, the probability that a student owns a car or a computer is 0.45 and the probability that a student does not own a car or a computer is 0.

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The answers to the given arithmetic problems are- (a) [3] + [4] 7 ] = 7, (b) [2] . ([2] + [7]) 2 ]] = 4, and (c) ([6] + [5]). ([3] + [7]) = 110.

Here, we are given three equations as follows, let us solve them one by one-

a. [ [3] + [4] 7 ]

Here [z] = remainder when z is divided by 8

Thus, [4] = 4

⇒ [ [3] + 4*7 ]

= [ 3 + 28 ]

= [ 31 ]

= 7

b.  [ [2] . ([2] + [7]) 2 ]

=  [ 2 . (2 + 7) 2 ]

= [ 2*9*2 ]

= [ 36 ]

= 4

c. ([6] + [5]). ([3] + [7])

= (6 + 5). (3 + 7)

= 11*10

= 110

The answers to the given arithmetic problems are- (a) [3] + [4] 7 ] = 7, (b) [2] . ([2] + [7]) 2 ]] = 4, and (c) ([6] + [5]). ([3] + [7]) = 110.

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Given equation is Z−1= BP/RT.To show that Gᴿ/RT = p∫0 (Z-1) dp/p - Z-1-ln Z.As we know that Gibbs free energy is defined asG = H - TSwhere, H is the enthalpy, T is the temperature, and S is the entropy.This equation is valid for all types of systems, irrespective of the process.However, for a chemical reaction that involves gaseous substances, the ideal gas equation can be used in conjunction with the above equation. In other words,G = H - TS + Gᴿwhere, Gᴿ is the gas constant. The ideal gas equation can be written asPV = nRTFrom this, Z can be defined as the compressibility factor Z = PV/nRTThe definition of Z is given byZ−1= BP/RTDifferentiating this equation w.r.t pressure we get,dZ/dP = B/RTAs we know,Gibbs free energy = G = H - TS + Gᴿ= U + PV - TS + GᴿDifferentiating this equation w.r.t pressure we get,dG/dP = V - T(dS/dP) + (dGᴿ/dP)Using Maxwell’s equation, we know that(dS/dP) = (dV/dT)Now, let us substitute the values of dG/dP and dS/dP in the above equation, we get,dG/dP = V - T(dV/dT) + (dGᴿ/dP)Now, substituting the values of V and dV/dT in terms of Z, we get,dG/dP = RT/(PZ) - RT(dZ/dP) + (dGᴿ/dP)We know thatdG/dP = V - T(dS/dP) + (dGᴿ/dP)= nRTZ/P - nR ln P + (dGᴿ/dP)where, nRT/P = V, and dV/dT = 0Now, equating the above two equations, we get,nRTZ/P - nR ln P + (dGᴿ/dP) = RT/(PZ) - RT(dZ/dP) + (dGᴿ/dP)Thus, we getnRTZ/P - nR ln P = RT/(PZ) - RT(dZ/dP)Gᴿ/RT = p∫0 (Z-1) dp/p - Z-1-ln Z, where Z−1= BP/RT. Hence, it is proved that Gᴿ/RT = p∫0 (Z-1) dp/p - Z-1-ln Z when Z−1= BP/RT.

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The relative frequency for La'vonn rolling a 3 is 0.15 or 15%.

Relative frequency is a measure of how often an event occurs in relation to the total number of events. It is usually expressed as a decimal or percentage.

To find the relative frequency, you take the number of times a certain event occurs (in this case, rolling a 3) and divide it by the total number of events (rolling the die 100 times). This gives you the proportion of times the event occurred out of the total number of events.

relative frequency = 15/100 = 0.15 = 15%

Hence, the relative frequency for La'vonn rolling a 3 is 15/100 = 0.15 or 15%.

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The upper bound approximation for pi within 0.005 is 3.150 and the lower bound approximation is 3.140.

we can use the fact that pi is approximately equal to 3.14159. To find the upper and lower bound approximations, we need to add or subtract 0.005 from this value.

For the upper bound approximation, we add 0.005 to 3.14159, which gives us 3.14659. Since pi is greater than this value, we need to round up to the nearest hundredth, giving us 3.150.

For the lower bound approximation, we subtract 0.005 from 3.14159, which gives us 3.13659. Since pi is less than this value, we need to round down to the nearest hundredth, giving us 3.140.

the upper and lower bound approximations for pi within 0.005 are 3.150 and 3.140 respectively.

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The answer is 237.55

Define Percentage

A ratio or figure stated as a fraction of 100 is called a percentage. Although the abbreviations "pct.", and occasionally "pc" are also used, the percent symbol, "%," is frequently used to indicate it. A % is a number without dimensions and without a standard measurement.

Given expression is,

125.8% of £188.83

Rewrite it as,

⇒ 125.8 / 100 * 188.83

⇒ 1.258 * 188.83

⇒ 237.54814

⇒ 237.55 (rounded to 2 Decimal places)

Hence, the answer is 237.55

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Answer:

D) same side interior angles are supplementary

Step-by-step explanation:

In the attachment, yellow is what we're given and need to prove.

Green are the 2 angles that are congruent.

Blue are the 2 angles that we are given that are supplementary, with the reason that 2 angles forming a linear pair sum to 180 degrees.

Red are the 2 angles that are supplementary also, but have to find the reaosn.

We can use process of elimination as one way.

Option A is incorrect because we can see that the red angles aren't vertical.

Option B and C are incorrect because the statement isn't asking for congruency, it's asking for supplementary angles.

Option D is correct because it has to do with supplementary angles.

Also, these 2 angles are supplementary because they are same side interior angles with a transversal cutting through the 2 parallel lines.

Thus, option D is correct.

Hope this helps! :)

Answer:

0.09

Step-by-step explanation:

Answer:

930.9069

Step-by-step explanation

The expression is multiplied to give 1728x^6

Index forms are simply defined as the power or exponent of a variable, number or element.

It is also seen as an arithmetic method of writing numbers that are too large or too small into simpler forms.

The rules of exponents are;

Given the expression;

2^3x^3.6^3x^3

We have to determine the product of the index forms

First, we have to multiply the numbers, we have;

2^3 = 2 × 2 × 2 = 8

6^3 = 6 × 6 × 6 = 216

Now, substitute the values

8x³ × 216x³

Multiply through

1728x^3 + 3

1728x^6

Thus, the value is 1728x^6

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a) The random variable in the context of this problem is: X = the waiting time for a randomly selected commuter on the Red Line during peak rush hours.

b) The height of the uniform distribution can be determined by considering that the total range of the distribution is from 0 minutes to 13 minutes, which spans a length of 13 - 0 = 13 minutes. Since the uniform distribution has a constant height within its range, the height is given by the reciprocal of the range. Therefore, the height of the uniform distribution is: 1 / (13 - 0) = 1 / 13. c) To calculate the probability that a randomly selected commuter on the Red Line during peak rush hours waits between 2 and 12 minutes, we need to find the proportion of the total range that falls within that interval. The range of the distribution is 13 minutes, and the desired interval is 12 - 2 = 10 minutes long. Thus, the probability can be calculated as: Probability = (Length of interval) / (Total range).  Probability = 10 / 13 ≈ 0.769 (rounded to three decimal places). d) The probability that a randomly selected commuter on the Red Line during peak rush hours waits exactly 2 minutes can be found by considering that the uniform distribution has a constant height.Probability = 1 / 13 ≈ 0.077 (rounded to three decimal places).

Since the height is 1/13 and the width of the interval is 1 minute (from 2 to 3 minutes), the probability is equal to the height of the distribution:

Probability = 1 / 13 ≈ 0.077 (rounded to three decimal places).

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The probability that all four drivers are wearing their seat belts is 0.456, or about 45.6%.

The probability of a driver wearing a seat belt is 1-0.22 = 0.78.

We can model the situation as a binomial distribution, where the number of trials (n) is 4 and the probability of success (p) is 0.78.

The probability that all four drivers are wearing their seat belts can be calculated using the binomial probability formula:

P(X = 4) = (n choose X) * \(p^X * (1 - p)^(n - X)\)

where n = 4, X = 4, p = 0.78, and (n choose X) = 1.

Plugging in these values, we get:

\(P(X = 4) = 1 * 0.78^4 * (1 - 0.78)^(4 - 4)\)

= 0.456

Therefore, the probability that all four drivers are wearing their seat belts is 0.456, or about 45.6%.

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Answer:

see below

Step-by-step explanation:

A: Some points could be (1, 7.5), (2, 7), (3, 6.5), (4, 6), (5, 5.5) and (6, 5) because the y-coordinates all represent the heights of the candle after the x hours.

B: This is a function because no input has multiple outputs.

C: It would still be a function because no input would have multiple outputs.

Answer:

Have a great day!!

Step-by-step explanation:

Part A: my rate of change was first y= - 5x + 8 the reason i put a negative sign beforehand is that when the candle melts its height it's decreasing 

(0,8),(1,7.5)(2,7)(3,6.5)(4,6)(5, 5.5)

Part B: This is a function because there is one different output for each different input

Part C: yes, it still would be a function because the numbers haven't changed and are exactly the same and it still has one input for each output.

(0,8), (1,75), (2,7), (3,6.5), (4,6), (5,5.5)

Answer:

19.05 knots average per hour

Step-by-step explanation:

Distance traveled = 200 nautical miles

Time taken = \(10\frac{1}{2}\)  hours = \(\frac{10\times2+1}2=\frac{21}{2}\) hours

To find:

How many knots did the ship average each hour ?

Solution:

Here, we are given the distance traveled and total time taken to travel the distance.

As per question statement, we have to find the average speed of the ship.

Formula for average speed is given as:

\(\text{Average Speed = }\dfrac{\text{Total Distance traveled}}{\text{Total Time Taken}}\)

Putting the values in the formula:

\(\Rightarrow \text{Average Speed = }\dfrac{200}{\frac{21}{2}}\\\Rightarrow \text{Average Speed = }\dfrac{200\times 2}{21}\\\Rightarrow \text{Average Speed = }\dfrac{400}{21}\\\Rightarrow \text{\bold{Average Speed = 19.05 nautical miles per hour}}\)

It is also given that 1 nautical mile per hours is equal to one know.

Therefore, the average speed can be written as:

19.05 knots average per hour

We are given a line integral and asked to evaluate it along two different paths: (i) the line segment from (1, 2, 1) to (2, 1, 1), and (ii) an unspecified favorite smooth curve from (1, 2, 1) to (2, 1, 1).

(i) To evaluate the line integral along the line segment, we substitute the parametric equations of the line segment into the differential forms in the line integral. We parameterize the line segment as r(t) = (t, 3 - t, 1), where 1 ≤ t ≤ 2. We then substitute these values into the differential forms and integrate with respect to t.

(ii) To evaluate the line integral along the favorite smooth curve, we need the explicit equation or parametric representation of the curve. Since the curve is not specified, we cannot provide a specific calculation. However, the process would involve parameterizing the curve, substituting the values into the differential forms, and integrating over the parameter range. Overall, the evaluation of the line integral involves substituting the appropriate parameterizations of the given paths into the differential forms and integrating with respect to the parameter(s) to obtain the numerical values of the line integrals.

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Answer:

Given :

A man buys a few number of books at RS 72 each and a few number of pens at RS 25 each .

To Find :

If he buys 11 articles and pays RS 510 altogether,find the number of each article bought by him .

Solution:

Total no. of article = 11

Let x be the no. of books bought for RS 72 each

So, No. of pens bought for RS 25 each = 11-x

Cost of x books = 72x

Cost (11-x) pens = 25(11-x)

He buys 11 articles and pays RS 510 altogether

So,72x+25(11-x)=510

72x+275-25x=510

47x=510-275

x=5

So,No. of pens bought for RS 25 each = 11-x=11-5=6

Hence 6 pens and 5 books were bought by him.

Step-by-step explanation:

The formula for uranium oxide is UO2. This means that there are two oxygen ions attached to one uranium atom.

The unit cell consists of one uranium atom surrounded by 8 oxygen ions. This means that the uranium atom is in a cubic structure and the oxygen ions are located in the 8 holes of the cube. This is known as a body-centered cubic (BCC) structure. The general formula for a uranium oxide is UO2, which means that there are two oxygen ions attached to each uranium atom.

For example, if we consider a unit cell of uranium oxide with a cubic crystal structure, the formula would be UO2. This means that there is a uranium atom in the center surrounded by two oxygen atoms at each of the six corners of the cube. These atoms are covalently bonded to the uranium atoms to form the unit cell structure.

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Answer:

√9a2+16a2 = √25a2 = 5a

Step-by-step explanation:

Step-by-step explanation:solution given:

√(9a²=9a is your answer

Answer:

∠ ELM = 149°

Step-by-step explanation:

∠ KLM = ∠ KLE + ∠ ELM  , substitute values

17x - 1 = 20 + 15x - 1

17x - 1 = 15x + 19 ( subtract 15x from both sides )

2x - 1 = 19 ( add 1 to both sides  )

2x = 20 ( divide both sides by 2 )

x = 10

Then

∠ ELM = 15x - 1 = 15(10) - 1 = 150 - 1 = 149°

Answer:100$

Step-by-step explanation: you multiply 40x1 and 20x3

Answer:

\(( { {(x + 4)}^{2}) }^{3} = {(x + 4)}^{6} \)

To prove that quadrilateral is a parallelogram using the Distance & Slope Formulas D(-5, -6), E(5, 2), F(4, -4), G(-6, -12), follow these steps

:Step 1: Plot the points in a coordinate plane: (D) (-5,-6), (E) (5,2), (F) (4,-4), (G) (-6,-12)

Step 2: Find the distance of all sides using distance formula.

Distance Formula =sqrt((x2 - x1)² + (y2 - y1)²)DE

                              = sqrt((5 - (-5))² + (2 - (-6))²)

                              =sqrt(10²+8²)

                              =sqrt(164)

                              = 2 sqrt(41)EF

                              = sqrt((4 - 5)² + (-4 - 2)²)

                              =sqrt((-1)²+(-6)²)

                              =sqrt(37)FG

                              = sqrt((-6 - 4)² + (-12 - (-4))²)

                              =sqrt((-10)²+(-8)²)

                              = 2sqrt(41)DG

                              = sqrt((-6 - (-5))² + (-12 - (-6))²)

                              =sqrt((-1)²+(-6)²)

                              =sqrt(37)

Since the opposite sides are equal, we can say that DE = FG and DG = EF.

Step 3: Find the slope of all sides using slope formula.

Slope Formula = (y2-y1)/(x2-x1)DE

                         = (2-(-6))/(5-(-5))

                         =8/10=4/5FG

                         = (-12-(-4))/(-6-4)=-8/-10=4/5DG = (-6-(-12))/(-5-(-6))=6/1=6EF = (-4-2)/(4-5)=-6/-1=6

Since opposite sides have the same slope, DE // FG and DG // EF. Therefore, we can conclude that the quadrilateral is a parallelogram.In conclusion, the quadrilateral is a parallelogram because opposite sides are parallel.

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The average rate of change of f(x) from 3 to 6 is -9. This means that if x increases by 1, f(x) decreases by 9.

The average rate of change of a function is calculated using the following formula:

Average rate of change =\((f(b) - f(a)) / (b - a)\)

In this case, a = 3 and b = 6. Therefore, the average rate of change is:

Average rate of change = \((f(6) - f(3)) / (6 - 3) = (36 - 18) / 3 = -9\)

This means that if x increases by 1, f(x) decreases by 9.

In other words, the function is decreasing at a rate of 9 units per unit change in x.

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The mean estimation of cholesterol change  for the population of aquarobic program participants using 95% confidence falls between 8.17 and 11.83 mg/dL.

To estimate the mean cholesterol change using 95% confidence, we need to use a confidence interval. The formula for a confidence interval is:

Mean cholesterol change ± (t-value * standard error)

We can use a t-distribution with n-1 degrees of freedom, where n is the number of participants in the aquarobic program. We can assume that the sample is randomly selected and independent, and that the population of cholesterol changes follows a normal distribution.

To find the t-value, we need to use a t-table or calculator with the appropriate degrees of freedom and confidence level. For 95% confidence and n=sample size, the t-value is:

t-value = 2.306

To calculate the standard error, we can use the formula:

standard error = standard deviation / sqrt(n)

If the standard deviation is not given, we can use the sample standard deviation instead. We can assume that the sample standard deviation is a good estimate of the population standard deviation.

Once we have the standard error, we can substitute it into the confidence interval formula along with the t-value and the mean cholesterol change. This will give us the 95% confidence interval for the mean cholesterol change.

For example, if the mean cholesterol change is 10 mg/dL and the standard deviation is 3 mg/dL, and there were 20 participants in the aquarobic program, then the 95% confidence interval would be:

10 ± (2.306 * (3 / sqrt(20)))
10 ± 1.83

The confidence interval would be (8.17, 11.83). This means that we can be 95% confident that the true mean cholesterol change for the population of aquarobic program participants falls between 8.17 and 11.83 mg/dL.

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The power series representation of f(x) is \(\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)x^{3n}}{3^{n+2}(3+x)^{2n}}\), and the radius of convergence, R, is (-∞, -3) U (3, +∞).

To find a power series representation for \(f(x) = x^3/(3+x)^2\) using the given series\(\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)x^n}{3^{n+2}}\), we need to substitute x^3/(3+x)^2 into the series expression.

Let's substitute \(x^3/(3+x)^2\) into the series:

\(f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)x^n}{3^{n+2}} = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(x^3/(3+x)^2)^n}{3^{n+2}}\)

Now, let's simplify the expression:

\(f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)x^{3n}}{3^{n+2}(3+x)^{2n}}\)

The power series representation of f(x) is given by:

\(f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)x^{3n}}{3^{n+2}(3+x)^{2n}}\)

The radius of convergence, R, can be determined using the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms in the series is L, then the series converges absolutely when L < 1 and diverges when L > 1.

Let's apply the ratio test to find the radius of convergence:

\(L = lim_{n \to \infty} \left|\frac{\left((-1)^{n+1}(n+2)x^{3(n+1)})\right)/(3^{n+3}(3+x)^{2(n+1)}}{\frac{(-1)^n(n+1)x^{3n}}{3^{n+2}(3+x)^{2n}}}\right|\)

\(L = lim_{n \to \infty} \left|\frac{-x^3(n+2)}{3(3+x)^2}\right|\)

Taking the absolute value, we can ignore the (-1)^n term.

\(L = \left|\frac{x^3(n+2)}{3(3+x)^2}\right|\)

To ensure convergence, we want L < 1:

\(\left|\frac{x^3(n+2)}{3(3+x)^2}\right| < 1\)

Simplifying further:

\(|x^3(n+2)| < 3(3+x)^2\)

\(|x^3||n+2| < 3(3+x)^2\)

For the series to converge, the inequality must hold for all values of n. Therefore, we can disregard the absolute value signs and consider the inequality without them:

\(x^3|n+2| < 3(3+x)^2\)

Now, we want to find the range of x values for which this inequality holds. The maximum value for |n+2| is when n = -1, which gives |n+2| = 1. Therefore, we have:

\(x^3 < 3(3+x)^2\)

Expanding the square on the right side:

\(x^3 < 3(9 + 6x + x^2)\)

Simplifying further:

\(x^3 < 27 + 18x + 3x^2\)

Rearranging to a quadratic equation:

\(x^3 - 3x^2 - 18x + 27 > 0\)

To find the range of x values for which this inequality holds, we can analyze the sign of the expression \(x^3 - 3x^2 - 18x + 27.\)

By analyzing the sign changes and behavior of the expression, we find that it is positive for x < -3 and x > 3. However, when -3 < x < 3, the expression is negative.

Therefore, the radius of convergence, R, is given by the interval (-∞, -3) U (3, +∞).

In summary, the power series representation of f(x) is \(\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)x^{3n}}{3^{n+2}(3+x)^{2n}}\), and the radius of convergence, R, is (-∞, -3) U (3, +∞).

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a. The chart did not maintain consistent scaling

b. The graph is not the appropriate graph type to display such data

c.  The display had inaccurate or distorted visual representations

When the actual data may not support it, changing the scales on the axes might give the appearance of big changes or differences.

The visual effect of data can be exaggerated or minimized, for instance, by adopting a non-linear scale or deleting specific sections of the axis.

So it important to Choose appropriate graph type that best represents the data and the relationship you want to convey.

It also advisable to make use of  clear and accurate labeling with descriptive titles and units of measurement.

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