EXAMPLE 2 The arc of the parabola y = 3x2 from (5, 75) to (10, 300) is rotated about the y-axis. Find the area of the resulting surface. SOLUTION 1 Using y = 3x2 and dy dx = we have, from this formula, S = 2πx ds = 10 2πx 1 + dy dx 2 dx 5 = 2π 10 x 1 + 36x2 dx 5 . Substituting u = 1 + 36x2, we have du = dx. Remembering to change the limits of integration, we have S = π 36 3601 u du 901 = π 36 3601 901 = . SOLUTION 2 Using x = y 3 and dx dy = we have S = 2πx ds = 300 2πx 1 + dx dy 2 dy 75 = 2π 300 y 3 1 + 1 12y dy 75 = π 3 300 12y + 1 dy 75 = π 36 3601 u du 901 (where u = 1 + 12y) = (as in Solution 1)

Answer:   It will take 4 days to complete the whole trip.

Step-by-step explanation:

Given : Karen is planning to drive 848 miles on a road trip.

The number of miles she travels in a day = 212 miles

Then, the number of days taken to complete the whole trip will be [Divide 848 by 212]:_

[tex]\dfrac{848}{212}=4[/tex]

Hence, it will take 4 days to complete the whole trip.

Final answer:

To find the number of days Karen needs to complete her 848-mile road trip at a rate of 212 miles per day, divide the total miles by the daily miles, which results in 4 days.

Explanation:

The student's question asks: Karen is planning to drive 848 miles on a road trip. If she drives 212 miles a day, how many days will it take to complete the trip? This is a simple division problem in mathematics. To find out how many days it will take for Karen to complete the trip, you divide the total miles of the trip (848 miles) by the number of miles she can drive in a day (212 miles/day).

So the calculation would be: 848 miles ÷ 212 miles/day = 4 day

Therefore, it will take Karen 4 days to complete her road trip if she drives 212 miles each day.

Answer: There are 196.85 cm in 500 inches.

Step-by-step explanation:

Since we have given that

1 inch = 2.54 centimeters

We need to find the number of centimeters are in 500 inches.

Since 1 inch = 2.54 centimeters

1 inch = [tex]\dfrac{1}{2.54}\ cm[/tex]

500 inches = [tex]\dfrac{500}{2.54}=196.85\ cm[/tex]

Hence, There are 196.85 cm in 500 inches.

To convert 500 inches to centimeters, use the conversion factor 2.54 cm = 1 inch. Cross-multiply and solve for x to find that there are approximately 1270 centimeters in 500 inches.

To convert inches to centimeters, we use the conversion factor 2.54 cm = 1 inch. In this case, we have 500 inches, so we can set up the following proportion:

We can cross-multiply and solve for x:

Calculating x, we find that there are approximately 1270 centimeters in 500 inches.

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Answer:

Option d - 204 m

Step-by-step explanation:

Given : The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 kg/m³.

To find : The height of the building ?

Solution :

We have given atmospheric pressures,

[tex]P_{\text{top}}=96\ kPa[/tex]

[tex]P_{\text{bottom}}=98\ kPa[/tex]

The density of air is 1.0 kg/m³ i.e. [tex]\rho_a=1\ kg/m^3[/tex]

Atmospheric pressure reduces with altitude,

The height of the building is given by formula,

[tex]H=\frac{\triangle P}{\rho_a\times g}[/tex]

[tex]H=\frac{P_{\text{bottom}}-P_{\text{top}}}{\rho_a\times g}[/tex]

[tex]H=\frac{(98-96)\times 10^3}{1\times 9.8}[/tex]

[tex]H=\frac{2000}{9.8}[/tex]

[tex]H=204\ m[/tex]

Therefore, Option d is correct.

The height of the building is 204 meter.

Using the barometric pressure formula and the given atmospheric pressures at the top and bottom of the building, the height is calculated to be approximately 204 meters, which matches option (d).

To calculate the height of the building using the difference in atmospheric pressure at the top and bottom, we can use the barometric pressure formula P = h ρ g, where P is the pressure, h is the height, ρ is the density of the fluid (or air in this case), and g is the acceleration due to gravity (approximately 9.8 m/s²).

The difference in pressure between the two points can be used to solve for h.

Given:

Difference in atmospheric pressure ΔP = 98.0 kPa - 96.0 kPa = 2.0 kPa

Density of air ρ = 1.0 kg/m3

Acceleration due to gravity g = 9.8 m/s²

We rearrange the formula to solve for h: h = ΔP / ( ρg)

h = (2.0 kPa) / (1.0 kg/m³ · 9.8 m/s²)

h = (2000 Pa) / (9.8 N/kg)

h = 204.08 m

The height of the building is therefore approximately 204 meters, making option (d) the correct answer.

Answer:

Step-by-step explanation:

Let's remember the equation of a line:

y = mx + b

m: slope (we know it's 3)

b: the y-intercept

So far, we have y = 3x + b, now we need to find b.

Replacing y and x for the given points (-3,-5):

-5 = 3*(-3) + b

-5 = -9 + b

b = -5 + 9

b = 4

The equation of the line that passes through the point (-3,-5) with a slope of 3 is y = 3x + 4

Answer:

y = 3x + 4

Step-by-step explanation:

Answer:

angle k:75

angle g:15

Step-by-step explanation:

Answer:

15,384 such tablets may be prepared from 5 kg of aspirin

Step-by-step explanation:

The problem states that aspirin tablets generally contain 325 mg of aspirin. And asks how many such tablets may be prepared from 5 kg of aspirin.

Since the problem measures the weight of a tablet in kg, the first step is the conversion of 325mg to kg.

Each kg has 1,000,000mg. So

1kg - 1,000,000mg

xkg - 325mg.

1,000,000x = 325

[tex]x = \frac{325}{1,000,000}[/tex]

x = 0.000325kg

Each tablet generally contains 0.000325kg of aspirin. How many such tablets may be prepared from 5 kg of aspirin?

1 tablet - 0.000325kg

x tablets - 5kg

0.000325x = 5

[tex]x = \frac{5}{0.000325}[/tex]

x = 15,384 tablets

15,384 such tablets may be prepared from 5 kg of aspirin

Around 15,385 tablets of aspirin can be created from 5 kilograms of aspirin, using the assumption that one tablet typically contains 325 milligrams of aspirin.

This question requires a basic understanding of the conversion from kilograms to milligrams. 5 kilograms of aspirin is equal to 5,000,000 milligrams (as 1 kilogram = 1,000,000 milligrams). So, if one aspirin tablet contains 325 mg of aspirin, we can calculate the number of tablets from 5kg, or 5,000,000 mg, of aspirin by simply dividing the total milligrams of aspirin by the milligrams per tablet.

To calculate:
Number of tablets = Total mass / Mass per tablet
= 5,000,000 mg / 325 mg/tablet

So, roughly 15,385 tablets of aspirin can be produced from 5 kg of aspirin.

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Answer:

  (4/7) ÷ (7/5) = 20/49

Step-by-step explanation:

You show a sum, not a quotient.

Perhaps you want the quotient ...

  (4/7) ÷ (7/5) = (4/7)×(5/7) = (4·5)/(7·7) = 20/49

The explanatory variable is the web page design. The response variables are the amount spent by the visitor and the amount of time visiting the site. Confounding could be an issue if other factors are not accounted for.

(a) The explanatory variable in this study is the web page design. It is a qualitative variable because it represents different design options.

(b) The two response variables are the amount spent by the visitor and the amount of time visiting the site. The amount spent is a quantitative variable, while the amount of time is also quantitative.

(c) Confounding could be an issue in this study if there are other factors that could affect the response variables and are not accounted for. For example, if there are differences in the age groups of the visitors using mobile and desktop platforms, their preferences may affect the response variables.

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Answer:

a) 18 cm

b) 18

c) 3

The Earl's height is unusual because the  z score does not lies in the given range of usual i.e -2 and 2

Step-by-step explanation:

Given:

Mean height, μ = 174 cm

Standard deviation = 6 cm

height of Earl, x = 192 cm

a) The positive difference between Earl height and the mean = x - μ

= 192 - 174 = 18 cm

b) standard deviations is 18

c) Now,

the z score is calculated as:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

or

[tex]z=\frac{192-174}{6}[/tex]

or

z = 3

The Earl's height is unusual because the  z score does not lies in the given range of usual i.e -2 and 2

Final answer:

Earl's height is 18 cm above the mean, which is 3 standard deviations from the mean, resulting in a z-score of 3. Considering the usual z-score range of -2 to 2, Earl's height is hence considered unusual.

Explanation:

The problem at hand involves understanding the concepts of statistics, particularly regarding mean, standard deviation, and z-scores in the context of normal distribution.

a. Positive Difference Between Earl's Height and the Mean

The positive difference between Earl's height and the mean is simply calculated by subtracting the mean from Earl's height. If Earl's height is 192 cm and the mean height is 174 cm, the difference is:

192 cm - 174 cm = 18 cm

b. Standard Deviations from the Mean

The number of standard deviations from the mean is found by dividing the difference by the standard deviation. Since the standard deviation is 6 cm,
18 cm / 6 cm = 3 standard deviations

c. Z-score Conversion

The z-score is calculated using the formula:

z = (Earl's height - mean) / standard deviation

z = (192 cm - 174 cm) / 6 cm = 3

d. Usuality of Earl's Height

Since "usual" heights translate to z-scores between -2 and 2, a z-score of 3 indicates that Earl's height is unusual.

Answer:

(a) PROPORTION.

Step-by-step explanation:

A relative frequency histogram has bars whose height is equal to either the proportion of cases that are between the upper and lower bounds of the bar.

The relative frequency in a relative frequency histogram refers to PROPORTION.

A relative frequency histogram is a graph that has the same shape and the same horizontal scale as the corresponding frequency histogram. The difference is that the vertical scale measures the relative frequencies (percentages or proportions).

Hence, option (a) is right option.

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

[tex]\frac{dC}{dt}=Ci*Qi-Co*Qo[/tex]

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

[tex]\frac{dC}{dt}=-Co*Qo[/tex]

Rearranging the equation, it becomes

[tex]\frac{dC}{C}=-Qo*dt[/tex]

Integrating both sides

[tex]\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}[/tex]

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

[tex]C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\[/tex]

The final equation for the concentration of salt at any given time is

[tex]C=exp^{-3*t-0.693}[/tex]

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

[tex]C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}=  -\frac{-0.266}{3}=0.088[/tex]

Answer:

After 7.84 the tank be half empty. The water remain in the tank after 5 days is 198.401 L.

Step-by-step explanation:

Consider the provided information.

It is given that a small hole in its base at a rate proportional to the square root of the volume of water remaining. The tank initially contains 300 liters and 22 liters leak out during the first day.

The rate of water leak can be written as:

[tex]\frac{dV}{dt}\propto \sqrt{V}[/tex]

Let k be the constant of proportionality.

[tex]\frac{dV}{dt}=k \sqrt{V}[/tex]

Integrate both the sides as shown:

[tex]\frac{dV}{\sqrt{V}}=k dt\\\int\frac{dV}{\sqrt{V}}=\int k dt\\2\sqrt{V} =kt+c[/tex]

Since for t=0 the volume was 300.

[tex]2\sqrt{300} =k(0)+c\\20\sqrt{3} =c\\c=34.641[/tex]

Now substitute the value of c in above equation.

[tex]2\sqrt{V} =kt+34.641[/tex]

22 liters leak out during the first day, thus now the remaining volume is 300-22=278 liters.

[tex]2\sqrt{278} =k(1)+34.641\\33.347 =k+34.641\\k=33.347 -34.641\\k=-1.294[/tex]

Thus, the required equation is:[tex]2\sqrt{V} =-1.294t+34.641[/tex]

Part (A) When will the tank be half empty.

Substitute v=150 liters for half empty in above equation.

[tex]2\sqrt{150} =-1.294t+34.641[/tex]

[tex]24.495 =-1.294t+34.641[/tex]

[tex]-10.146 =-1.294t[/tex]

[tex]t=7.84[/tex]

Hence, after 7.84 the tank be half empty.

Part (B) How much water will remain in the tank after 5 days.

Substitute the value of t=5 in [tex]2\sqrt{V} =-1.294t+34.641[/tex]

[tex]2\sqrt{V} =-1.294(5)+34.641[/tex]

[tex]2\sqrt{V} =28.171[/tex]

[tex]\sqrt{V} =14.0855[/tex]

[tex]V =198.401[/tex]

Hence, the water remain in the tank after 5 days is 198.401 L.

The correct answer is A) The tank will be half empty in 16 days, B) The remaining volume after 5 days will be 198 L.

A) To find when the tank will be half empty, we need to solve the differential equation that models the rate of change of the volume of water in the tank.

Let V(t) be the volume of water remaining in the tank at time t.

The rate of change of the volume is proportional to the square root of the volume:

dV/dt = -k√V

where k is a constant that can be determined from the given information.

We know that V(0) = 300 L and V(1) = 300 - 22 = 278 L.

Substituting these values, we get:

k = 22 / √300 = 4

Solving the differential equation with the initial condition V(0) = 300, we get:

[tex]V(t) = 300^_{(1/2)}$-2t^_2[/tex]

Setting V(t) = 150 L (half of the initial volume), we get:

t = 16 days

B) To find the volume remaining after 5 days, we substitute t = 5 in the solution:

[tex]V(5) = (300^_(1/2)} - 2(5))^2 = (\sqrt{300} - 10)^2 = 198 L[/tex]

The rate of change of the volume is proportional to the square root of the volume, which leads to a separable differential equation. By using the given information to determine the constant of proportionality, we can solve the differential equation and find the time when the volume is halved. Substituting the desired time into the solution gives the remaining volume after that time.

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

[tex]x'=\frac{dx}{dt}[/tex]

a)  x’=t–sin(t),  x(0)=1

[tex]dx=(t-sint)dt[/tex]

Apply integral both sides:

[tex]\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k[/tex]

where k is a constant due to integration. With x(0)=1, substitute:

[tex]1=0+cos0+k\\\\1=1+k\\k=0[/tex]

Finally:

[tex]x=\frac{t^2}{2} +cos(t)[/tex]

b) x’+2x=4; x(0)=5

[tex]dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\[/tex]

Completing the integral:

[tex]-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}[/tex]

Solving the operator:

[tex]-\frac{1}{2}ln(4-2x)=t+k[/tex]

Using algebra, it becomes explicit:

[tex]x=2+ke^{-2t}[/tex]

With x(0)=5, substitute:

[tex]5=2+ke^{-2(0)}=2+k(1)\\\\k=3[/tex]

Finally:

[tex]x=2+3e^{-2t}[/tex]

c) x’’+4x=0; x(0)=0; x’(0)=1

Let [tex]x=e^{mt}[/tex] be the solution for the equation, then:

[tex]x'=me^{mt}\\x''=m^{2}e^{mt}[/tex]

Substituting these equations in c)

[tex]m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i[/tex]

This becomes the solution m=α±βi where α=0 and β=2

[tex]x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)[/tex]

Where A and B are constants. With x(0)=0; x’(0)=1:

[tex]x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}[/tex]

Finally:

[tex]x=\frac{1}{2} sin(2t)[/tex]

Answer:

E. t - 2t - 3t =32, t=8

Step-by-step explanation:

A. 2x + 5 (x-1)= 9, X= 2 it's ok because

[tex]2x+5(x-1)=9\\ 2x+5x-5=9\\ 7x=14\\ x=2[/tex]

B. p - 3(p-5)= 10, p= 2.5 it's ok as well because

[tex]p-3(p-5)=10\\ p-3p+15=10\\ -2p=-5\\ p=\frac{-5}{-2} \\ p=2.5[/tex]

C. 4 y +3 y=28, y=4 also it's ok because

[tex]4y+3y=28\\ 7y=28\\ y=\frac{28}{7} \\ y=4[/tex]

E. t - 2t - 3t =32, t=8 it's not ok because

[tex]t-2t-3t=32\\-4t=32\\t=\frac{32}{-4} \\t=-8[/tex]

a (-) is missing!

Final answer:

After reviewing the solutions provided by Mary, we can conclude that the incorrect solution is Option E, where the solution should be t = -8 instead of the provided answer t = 8.

Explanation:

Mary is reviewing her algebra quiz and needs to determine which one of her solutions is incorrect. We must go through each option and verify the results given.

Option A

2x + 5(x - 1) = 9, solution X = 2.

First, expand the equation: 2x + 5x - 5 = 9.

Then combine like terms: 7x - 5 = 9.

Next, add 5 to both sides: 7x = 14.

Finally, divide by 7: x = 2.

This solution is correct.

Option B

p - 3(p - 5) = 10, solution p = 2.5.

First, expand the equation: p - 3p + 15 = 10.

Then combine like terms: -2p + 15 = 10.

Next, subtract 15 from both sides: -2p = -5.

Finally, divide by -2: p = 2.5.

This solution is correct.

Option C

4y + 3y = 28, solution y = 4.

First, combine like terms: 7y = 28.

Then, divide by 7: y = 4.

This solution is correct.

Option E

t - 2t - 3t = 32, solution t = 8.

First, combine like terms: -4t = 32.

Next, divide by -4: t = -8.

This solution is incorrect because the given answer is t = 8. The correct answer should be t = -8.

Answer: There 64 possible sequences in which team A wins

Step-by-step explanation:

Hi!

The sequences in which team A wins, are the ones with at least 4 A's.

Sequences with 4, 5, 6 or 7 A's. To calculate how many of each type exist, we use the formula of combinations. If you select K objects from a set of N objects, there are C(N, K) possibilities, give by the formula:

[tex]C(N,K) = \frac{N!}{K! (N-K)!)}[/tex]

Then the total M number of sequences in which team A wins is:

[tex]M = C(7,7) + C(7,6) + C(7,5) + C(7,4) = 1 + 7 +21+35 = 64[/tex]

Answer:

36.

Step-by-step explanation:

We are asked to find the number of ways in which 3 girls can divide 10 pennies such that each must end up with at least one penny.

The selection can be done by selecting two dividing likes between the 10 pennies such that the set is divided into three parts.

Since each girl must have one penny, so no girl can have 0 penny. So the dividing like cannot be placed at end points, beginning and at the end. Therefore, we are left with 9 positions.  

Now we need to find number of ways to select two positions out of the 9 positions that is C(9,2).

[tex]_{2}^{9}\textrm{C}=\frac{9!}{7!*2!}=\frac{9*8*7!}{7!*2*1}=\frac{9*8}{2}=9*4=36[/tex]

Therefore, there are 36 ways to divide 10 pennies between 3 girls.

Answer:

10.27

Step-by-step explanation:

Find the number in the hundredth place  7  and look one place to the right for the rounding digit 1 . Round up if this number is greater than or equal to  5  and round down if it is less than  5 . And the answer is 10.27 which is rounded to the nearest hundredth.

Here, we are required to round the number 10.2719 to the nearest hundredth.

While considering place values of numbers,

Therefore, according to the question, the digit with occupies the hundredth position is 7.

As such, to round off the number 10.2719, the digit which is after the digit 7 is considered.

If the digit is less than 5, it is rounded to 0, and if greater or equal to 5, it is rounded to 1 and ultimately added to the preceding number.

In this case, the number is 1 and since 1 is less than 5, it is rounded to 0 and added to 7.

Ultimately, the number 10.2719 rounded to the nearest hundredth is;

10.27

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Answer:

The particular solution is [tex]y=\sin (\ln|x|)[/tex] .

Step-by-step explanation:

The given differential equation is

[tex]xy'=\sqrt {1-y^2}[/tex]

It can be written as

[tex]x\frac{dy}{dx}=\sqrt {1-y^2}[/tex]

Use variable separable method to solve the above equation.

[tex]\frac{dy}{\sqrt {1-y^2}}=\frac{1}{x}dx[/tex]

Integrate both sides.

[tex]\int \frac{dy}{\sqrt {1-y^2}}=\int \frac{1}{x}dx[/tex]

[tex]\sin^{-1} y=\ln|x|+C[/tex]            .... (1)

It is given that y(1)=0. It means y=0 at x=1.

[tex]\sin (0)=\ln|1|+C[/tex]

[tex]0=0+C[/tex]

[tex]0=C[/tex]

The value of constant is 0.

Substitute C=0 in equation (1) to find The required equation.

[tex]\sin^{-1} y=\ln|x|+0[/tex]

Taking sin both sides.

[tex]y=\sin (\ln|x|)[/tex]

Therefore the particular solution is [tex]y=\sin (\ln|x|)[/tex] .

To figure out how much E5 should be mixed with 6000 gal of E10 to make an E9 mixture, we need to use a weighted average equation: 0.10*6000 + 0.05*X = 0.09 * (6000 + X). Solving this equation will provide the required amount of E5 gasoline.

To solve this problem, we use a technique known as a weighted average. The weight is the number of gallons and the 'value' is the percentage of ethanol. We can formulate an equation using the principle that the sum of the ethanol in the initial gasolines will equal to the ethanol in the final mixture.

Let's denote by X the needed amount of E5 gasoline. Therefore, the total ethanol before mixing would be 0.10*6000 (from the E10 gasoline) + 0.05*X (from the E5 gasoline). After mixing, the total ethanol will be 0.09 (6000 + X).

Equating these two gives us 0.10*6000 + 0.05*X = 0.09 * (6000 + X). Solving this equation will give the needed amount of E5 gasoline.

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To create an E9 mixture, you need to mix 1500 gallons of E5 with 6000 gallons of E10. This is calculated using the volume percentages of ethanol in each mixture and solving a linear equation. The final mixture will contain 9% ethanol.

To solve the problem of finding out how much E5 should be mixed with 6000 gallons of E10 to create an E9 mixture, we need to set up an equation based on the percentage volumes of ethanol.

Let's denote the amount of E5 to be added as x (in gallons).

E10 contains 10% ethanol and 90% gasoline. Therefore, in 6000 gallons of E10, the amount of ethanol is:

E5 includes 5% ethanol and 95% gasoline. Therefore, in x gallons of E5, the amount of ethanol is:

We want to create an E9 mixture, which means the final mixture should contain 9% ethanol. The total volume of the final mixture will be:

The amount of ethanol in the final mixture can be represented as:

This amount should be 9% of the total final volume:

Solving for x:

Therefore, you need to mix 1500 gallons of E5 with 6000 gallons of E10 to create an E9 mixture.

Answer:

Bella can complete work alone in 120 minutes

Step-by-step explanation:

Ana and Bella together can  complete the shoveling in 30 minutes.

They can do a part of work in 1 minute = [tex]\frac{1}{30}[/tex]

Ana can do work alone in 40 minutes

She can do part of work in 1 minute = [tex]\frac{1}{40}[/tex]

So, Bella can do a part of work in 1  minute =  [tex]\frac{1}{30}-\frac{1}{40}[/tex]

                                                                        =  [tex]\frac{1}{120}[/tex]

Bella can do [tex]\frac{1}{120}[/tex] part of work in 1  minute

Bella can do whole work in minutes = 120

Hence Bella can complete work alone in 120 minutes

Answer:

([tex]\frac{9}{4},\frac{1}{4})[/tex]

Step-by-step explanation:

We are given that two lines

[tex]x_1-7x_2=5[/tex] and [tex]x_1-5x_2=1[/tex]

We have to find the intersection point of two lines

Let [tex]3x_1-7x_2=5[/tex] (equation 1)

[tex]x_1-5x_2=1[/tex] (Equation 2)

Multiply equation 2 by 3 then subtract  from  equation 1

[tex]-7x_2+15x_2=5-3[/tex]

[tex]8x_2=2[/tex]

[tex]x_2=\frac{2}{8}=\frac{1}{4}[/tex]

Substitute [tex] x_2=\frac{1}{4}[/tex] in the equation 1

Then, we get

[tex]3x_1-7\frac{1}{4}=5[/tex]

[tex]3x_1-\frac{7}{4}=5[/tex]

[tex]3x_1=5+\frac{7}{4}=\frac{20+7}{4}=\frac{27}{4}[/tex]

[tex]x_1=\frac{27}{4\times 3}=\frac{9}{4}[/tex]

Hence, the intersection point of two given lines is ([tex]\frac{9}{4},\frac{1}{4})[/tex]

The answer is

t dy / dt + dy / dt = te ^ y

I apply common factor                1/dt*(tdy+dy)=te^y

I pass "dt"                    tdy+dy=(te^y)*dt

I apply common factor                (t+1)*dy=(te^y)*dt

I pass "e^y"                         (1/e^y)dy=((t+1)*t)*dt

I apply integrals                ∫ (1/e^y)dy= ∫ (t^2+t)*dt

by property of integrals  ∫ (1/e^y)dy= ∫ (e^-y)dy

∫ (e^-y)dy= ∫ (t^2+t)*dt

I apply integrals

-e^-y=(t^3/3)+(t^2/2)+C

I apply natural logarithm to eliminate "e"

-ln (e^-y)=-ln(t^3/3)+(t^2/2)+C

y=ln(t^3/3)+(t^2/2)+C

The general solution of the differential equation is

[tex]y = - ln( - ln |t + 1| + c)[/tex]

How to find the particular solution.

Given differential equation

[tex]t \frac{dy}{dt} + \frac{dy}{dt} = t {e}^{y} [/tex]

we can rearrange it as:

[tex](t + 1) \frac{dy}{dt} = t {e}^{y} [/tex]

Now, we can separate variables:

[tex] \frac{dy}{t {e}^{y} } = \frac{dt}{t + 1} [/tex]

Integrating both sides

[tex]∫ \frac{1}{t {e}^{y} } dy = ∫ \frac{1}{t + 1} dt[/tex]

[tex] - {e}^{ - y} = ln |t + 1| + c[/tex]

[tex] {e}^{ - y} = ln |t + 1| + c[/tex]

[tex]y = - ln( - ln |t + 1| + c[/tex]

Where C1 is the constant of integration.

Answer: The value of t become 10000.

Step-by-step explanation:

Since we have given that

[tex]\dfrac{1}{2}\times 10t=50000[/tex]

We need to find the value of t from the above expression:

first we divide 10t by 2 to get 50000.

[tex]\dfrac{10t}{2}=50000\\\\5t=50000\\\\t=\dfrac{50000}{5}\\\\t=10000[/tex]

Hence, the value of t become 10000.

Answer:

The required answers are:

33,000,000,000

0.0000054

0.72

4.5

Step-by-step explanation:

Consider the provided information.

We can convert the scientific notation into long hand notation as shown:

If the exponent of 10 is a positive number then move the decimal point right as much as the exponent value.

If the exponent of 10 is a negative number then move the decimal point left as much as the exponent value.

Part (A) [tex]3.3\times 10^{10}[/tex]

Here the exponent of 10 is 10 which is a positive number, so move the decimal point right as shown:

[tex]3.3\times 10^{10}=33,000,000,000[/tex]

Part (B) [tex]5.4\times 10^{-6}[/tex]

Here the exponent of 10 is -6 which is a negative number, so move the decimal point left as shown:

[tex]5.4\times 10^{-6}=0.0000054[/tex]

Part (C) [tex]7.2\times 10^{-1}[/tex]

Here the exponent of 10 is -1 which is a negative number, so move the decimal point left as shown:

[tex]7.2\times 10^{-1}=0.72[/tex]

Part (D) [tex]4.5\times 10^0[/tex]

Here the exponent of 10 is 0. Use the property of exponent [tex]a^0=1[/tex]

[tex]4.5\times 10^0=4.5[/tex]

Answer:

The proof makes use of congruences as follows:

Step-by-step explanation:

We can prove this result using congruences module 3. First of all we shall show that

[tex]2^{2n-1}\equiv 2 \pmod{3}[/tex] for all [tex]n\in \mathbb{N}[/tex]. By induction we have

Then induction we proved that [tex]2^{2n-1}\equiv 2 \pmod{3}[/tex] for all [tex]n>1[/tex]. Then

[tex]2^{2n-1}+1\equiv 2+1\equiv 3\equiv 0 \pmod{3}[/tex]

From here we conclude that the expression [tex]2^{2n-1}+1[/tex] is divisible by 3.

Answer:

Probability of rolling a 2 and flipping a head will be [tex]\frac{1}{12}[/tex]    

Step-by-step explanation:

If we roll one die then probability to get any one side is [tex]\frac{1}{6}[/tex]

Therefore, probability to get 2 by rolling the die will be P(A) = [tex]\frac{1}{6}[/tex]

Now we flip a coin then getting head or tale probability is [tex]\frac{1}{2}[/tex]

Or probability to get head by flipping the coin P(B) = [tex]\frac{1}{2}[/tex]

Probability of happening both the events (rolling a 2 and flipping a head) will be denoted by

P(A∩B) = P(A)×P(B)

           = [tex]\frac{1}{6}\times \frac{1}{2}[/tex]

           = [tex]\frac{1}{12}[/tex]

Therefore, probability of rolling a 2 and flipping a head will be [tex]\frac{1}{12}[/tex]        

Answer:

The probability of rolling a 2 and flipping a head is [tex]\frac{1}{12}[/tex]

Step-by-step explanation:

Notice that rolling a die and flipping a coin are two independent events this means that the probability that one event occurs in no way affects the probability of the other event occurring. When we determine the probability of two independent events we multiply the probability of the first event by the probability of the second event.

[tex]P(X and \:Y) =P(X) \cdot P(Y)[/tex]

When you roll a die there six outcomes from 1 to 6 and when you flip a coin are two possible outcomes (heads or tails).

We know that the probability of an event is

[tex]P=\frac{the \:number \:of \:wanted\:outcomes}{the \:number \:of \:possible \:outcomes}[/tex]

So the probability of rolling a 2 and flipping a head is

[tex]P(2 \:and \:H) = P(2) \cdot P(H)\\P(2 \:and \:H) = \frac{1}{2} \cdot \frac{1}{6}\\P(2 \:and \:H) = \frac{1}{12}[/tex]

Answer:

[tex]-\frac{6}{37} + \frac{371}{37}i[/tex]

Step-by-step explanation:

We need to evaluate [tex]\frac{(5+6i)(5+6i)}{6+i}[/tex]

(5+6i)(5+6i) = (25 + 36i² + 60i) = (25 - 36 + 60i) = -11 + 60i

= [tex]\frac{-11+60i}{6+i}[/tex]

Now we rationalize the denominator.

Now, multiplying both the numerator and denominator by (6-i)

[tex]\frac{-66 + 11i + 360i - 60i^2}{36 - i^2} = \frac{-66 + 60  + 371i}{37} = \frac{-6 + 371i}{37}[/tex]

= [tex]-\frac{6}{37} + \frac{371}{37}i[/tex]

Formula used:

(a+b)² = a² + b² + 2ab

i² = -1

Answer:

  B.  1/10 the value of

Step-by-step explanation:

In our base-10 place-value number system, the place value of a number is multiplied by 10 when it moves 1 place to the left. It is multiplied by 1/10 when it moves 1 place to the right.

A digit has 1/10 the value of the same digit one place to its left.

Answer:    B

Step-by-step explanation:

Answer:

Extensive property

Step-by-step explanation:

The intensive properties does not depend on the amount of mass of the system or the size of the system, for example the density [tex]\rho[/tex] is a quantity that is already defined for the system, because the density of water is equal for a drop or for a pool of water.

In the case of extensive properties the value of them is proportional to the size or mass of the system. For example mass is an extensive property because depend on the amount of substance. Other example could be the enegy.

In the case of weigth is a quantity that depends on mass value, then weigth is an extensive property.

Answer:  (45.79, 51.21)

Step-by-step explanation:

Given : Significance level : [tex]\alpha: 1-0.99=0.01[/tex]

Sample size : n= 51 , which is a large sample (n>30), so we use z-test.

Critical value: [tex]z_{\alpha/2}=2.576[/tex]

Sample mean : [tex]\overline{x}= 48.5\text{ hours}[/tex]

Standard deviation : [tex]\sigma=7.5\text{ hours}[/tex]

The confidence interval for population means is given by :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

i.e. [tex]48.5\pm(2.576)\dfrac{7.5}{\sqrt{51}}[/tex]

i.e.[tex]48.5\pm2.70534112234\\\\\approx48.5\pm2.71\\\\=(48.5-2.71, 48.5+2.71)=(45.79, 51.21)[/tex]

Hence, 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week =  (45.79, 51.21)

Final answer:

To construct a 99% confidence interval for the standard deviation of firm's employees' work hours, follow the steps to calculate the interval between 6.26 and 9.08 hours.

Explanation:

Confidence Interval Calculation:

Calculate the degrees of freedom (df) using the formula df = n - 1. For a sample size of 51, df = 51 - 1 = 50.

Determine the critical values from the chi-squared distribution for a 99% confidence level with 50 degrees of freedom. These critical values are 30.984 and 73.361.

Next, calculate the confidence interval for the standard deviation using the formula CI = sqrt((n-1) ×s² / chi-squared upper) to sqrt((n-1) ×s² / chi-squared lower), which results in 6.26 to 9.08 hours.