M1Q6.) How many degrees should be used to represent convertables in the Pie Graph?

Answer:  Probability that either had Internet access or had cable television is 94%.

Step-by-step explanation:

Since we have given that

Probability of households had internet access P(I) = 85% = 0.85

Probability of households had cable television P(C) = 81% = 0.81

Probability of households had both Internet and cable television P(I ∩ C) = 72% = 0.72

We need to find the probability that either had Internet access or had cable television.

As we know the formula:

P(I ∪ C)=P(I) + P(C) - P(I ∩ C)

[tex]P(I\cup C)=0.85+0.81-0.72\\\\P(I\cup C)=0.94=94\%[/tex]

Hence, our required probability is 94%.

To determine the probability of a household having access to either the internet or cable television, the probabilities of both events must be added then subtract the probability of both occurring simultaneously. Doing this calculation, P(Internet) + P(Cable) - P(Both), we find the probability to be 94%.

In order to answer this question, we would need to use some principles of probability, specifically the rule for adding the probabilities of two mutually exclusive outcomes. But, because the outcomes in this case are not mutually exclusive (a household can have both Internet and cable television), we need to adjust our calculation by subtracting the probability of both outcomes occurring together.

The probability of having either Internet or cable television would be P(Internet) + P(Cable) - P(Both) which equals to [tex]85 + 81 - 72 = 94%.[/tex]Therefore, the probability of selecting a household that had either Internet access or cable television is 94%.

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Answer:Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)

Step-by-step explanation:

given f(x) = sin(x) + cos(x)

f(x) can be rewritten as [tex]\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} }  ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\[/tex]

Using these result in equation a we get

f(x) = [tex]\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)[/tex]

Now we know that for derivative with respect to dependent variable is positive for an increasing function

Differentiating b on both sides with respect to x we get

f '(x) = [tex]f '(x)=\sqrt{2}  \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0[/tex]

where x∈(0,2π)

we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]

Thus for cos(π/4+x)>0 we should have

1) π/4 + x < π/2  => x<π/4  => x∈[0,π/4]

2) π/4 + x > 3π/2  => x > 5π/4  => x∈[5π/4,2π]

from conditions 1 and 2 we have  x∈(0,π/4)∪(5π/4,2π)

Thus the function is decreasing in x∈(π/4,5π/4)

To find the intervals of increase and decrease for f(x) = sin x + cos x, we first find the derivative f'(x) = cos x - sin x. By setting the derivative equal to zero, we find the critical points at x = π/4 + kπ. By testing these intervals in the derivative, we can identify the intervals of increase and decrease.

The function given is f(x) = sin x + cos x, which is a combination of a sine and cosine function. To find the intervals where the function is increasing or decreasing, we need to find the derivative of the function first. The derivative of sin x is cos x, and the derivative of cos x is -sin x. So, the derivative of the function f(x) is f'(x) = cos x - sin x.

By setting the derivative equal to zero, cos x - sin x = 0, we can find the critical points where the function may change from increasing to decreasing or vice versa. The solutions to this equation are x = π/4 + kπ, where k is an integer.

Using these points, we can find the intervals of increase and decrease. For instance, if we test a number between 0 and π/4 in the derivative, we find that the function is increasing on the interval (0, π/4). Continuing this process for the rest of the intervals should provide all the intervals of increase and decrease for the function.

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The answer is:

[tex]t(s(5))=-77[/tex]

To solve the problem, first, we need to compose the functions, and then evaluate the obtained function. Composing function means evaluating a function into another function.

We have that:

[tex]f(g(x))=f(x)\circ g(x)[/tex]

From the statement we know the functions:

[tex]s(x)=-3x-4\\t(x)=4x-1[/tex]

We need to evaluate the function "s" into the function "t", so:

[tex]t(s(x))=4(-3x-4)-1\\\\t(s(x))=-12x-16-1=-12x-17[/tex]

Now, evaluating the function, we have:

[tex]t(s(5))=-12(5)-17=-60-17=-77[/tex]

Have a nice day!

Answer:

x = 12/7 or 1.7

Step-by-step explanation:

first, cross multiply to get 7x = 12. then, divide 12 by 7 to get 12/7, which can be simplified and rounded to 1.7.

Final answer:

To solve 7/4 = 3/x, use cross multiplication to get 7x = 12, then divide both sides by 7 to find x, which is approximately 1.7 when rounded to the nearest tenth.

Explanation:

To solve the equation 7/4 = 3/x, we can set up a proportion and use cross multiplication. Cross multiplication involves multiplying the numerator of one fraction by the denominator of the other fraction, and setting the products equal to each other. In this case, we multiply 7 by x and 4 by 3 to get the equation 7x = 12.

After cross multiplying, divide both sides of the equation by 7 to solve for x. Doing this, we find that x = 12/7. To convert this to a decimal and round to the nearest tenth, we can divide 12 by 7 using a calculator or long division, resulting in approximately 1.7.

Answer:

The lower limits (in 1000 ​cells/mu​L): 0,100, 200, 300, 400, 500, 600 .

The upper limits (in 1000 ​cells/mu​L): 99, 199, 299, 399, 499, 599, 699.

The class width(in 1000 ​cells/mu​L): 100.

Class midpoints (in 1000 ​cells/mu​L): 49.5, 149.5, 249.5, 349.5, 449.5, 549.5, 649.5.

Class boundaries (in 1000 ​cells/mu​L): -0.5, 99.5, 199.5, 299.5, 399.5, 499.5, 599.5, 699.5.

Individuals included in the summary: 155.

Step-by-step explanation:

For the lower class limit is needed the smallest value in each class: 0,100, 200, 300, 400, 500, 600.

For the upper-class limit is needed the biggest value in each class: 99, 199, 299, 399, 499, 599, 699.

The class width is the difference between the lower limit of one class and the lower limit of the previous class. For example, 200 is the lower limit of one class and the lower limit of the previous class is 100, so 200-100=100.

Class midpoints are the average of the limits of a class if the limits are 0 and 99 then:

[tex]Midpoint=\frac{0+99}{2}=49.5[/tex]

[tex]Midpoint=\frac{100+199}{2}=149.5[/tex]

[tex]Midpoint=\frac{200+299}{2}=249.5[/tex]

[tex]Midpoint=\frac{300+399}{2}=349.5[/tex]

[tex]Midpoint=\frac{400+499}{2}=449.5[/tex]

[tex]Midpoint=\frac{500+599}{2}=549.5[/tex]

[tex]Midpoint=\frac{600+699}{2}=649.5[/tex]

Class boundaries are the numbers than allow to separate each class, to find each one, first calculate the gap between each class (the lower limit of one class minus the upper limit of the previous one) and divide by 2:

100-99=1

1/2=0.5

Add this number to all the upper limit to find the upper boundaries:

[tex]99+0.5=99.5[/tex]

[tex]199+0.5=199.5[/tex]

[tex]299+0.5=299.5[/tex]

[tex]399+0.5=399.5[/tex]

[tex]499+0.5=499.5[/tex]

[tex]599+0.5=599.5[/tex]

[tex]699+0.5=699.5[/tex]

And  subtract this number (0.5) from the lower limit to find the lower boundaries:

[tex]0-0.5=-0.5[/tex]

[tex]100-0.5=99.5[/tex]

[tex]200-0.5=199.5[/tex]

[tex]300-0.5=299.5[/tex]

[tex]400-0.5=399.5[/tex]

[tex]500-0.5=499.5[/tex]

[tex]600-0.5=599.5[/tex]

The total of individuals is equal to the sum of all the frequencies of each class:

[tex]3+54+76+21+0+0+1= 155[/tex]

The lower class limits are 0, 100, 200, etc. The class width is 100. The summation of the frequencies, which equals the total number of individuals, is 155.

In the given data, the lower class limits are the smallest numbers in each class or group, which are 0, 100, 200, 300, 400, 500, and 600 (in 1000 cells/muL). The upper class limits, which are the highest numbers in each class, are 99, 199, 299, 399, 499, 599, and 699.

The class width, which is the difference between two consecutive lower class limits, is 100. The class midpoints can be calculated by adding the lower and upper limits of each class and divide by 2, yielding 49.5, 149.5, 249.5, etc. The class boundaries are the numbers that separate classes without leaving gaps: -0.5, 99.5, 199.5, etc.

The total number of individuals included in the summary is the sum of the frequencies, which equals 155 in this case.

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Answer:  1.029 tons/m³

Step-by-step explanation:

Density of sea water may varies for different temperature but at 25°C the density of sea water is taken as 1029 kg/m³.

Density of sea water is 1029 kg/m³

1 tons=1000 kg

1 kg=0.001 tons

Density of sea water=1029×0.001 tons/m³

                                  =1.029 tons/m³

hence the density will be 1.029 tons/m³

Answer:

According to the internet, the density of the water is 1023.6 kg/m^ 3, but when it is converted, it is 1.0273 tons/m^3.

Step-by-step explanation:

This is what I found, hope it helps.

Answer:

The correct option is B.

Step-by-step explanation:

Given information:  n(A) = 35, n(B) = 42, n(C) = 45, n(A∩B) = 8, n(A∩C) = 8, n(B∩C) = 6, and n(A∩B∩C) = 3.

We need to find the value of n(A∩(B∩C)')

Using venn diagram we get

n(A∩B∩C')=n(A∩B)-n(A∩B∩C)= 8-3 = 5

n(A∩B'∩C)=n(A∩C)-n(A∩B∩C)= 8-3 = 5

n(A'∩B∩C)=n(B∩C)-n(A∩B∩C)= 6-3 = 3

n(A∩(B∪C)')=n(A)-n(A∩B'∩C)-n(A∩B∩C')-n(A∩B∩C)

n(A∩(B∪C)')=35-5-5-3 = 22

The value of n(A∩(B∪C)') is 22. Therefore the correct option is B.

Answer: Hence, our required probability is [tex]\dfrac{1}{386206920}[/tex]

Step-by-step explanation:

Since we have given that

Numbers in a lottery = 60

Numbers to win the jackpot = 7 numbers

We need to find the probability to hit the jackpot:

So, our required probability is given by

[tex]P=\dfrac{^7C_7}{^{60}C_7}\\\\P=\dfrac{1}{386206920}[/tex]

This is a combination problem as we need to select 7 numbers irrespective of any arrangements.

Hence, our required probability is [tex]\dfrac{1}{386206920}[/tex

Answer: [tex]H_0:p\leq0.25[/tex]

[tex]H_a:p>0.25[/tex]

Step-by-step explanation:

Given : In the past , 25% of the town residents participated in the school board elections.

Let 'p' be the proportion of voters will participated in the school board elections.

Claim : [tex]p>0.25[/tex]

We know that the null hypothesis has equal sign.

Therefore , the null hypothesis for the given situation will be opposite to the given claim will be :-

[tex]H_0:p\leq0.25[/tex]

And the alternative hypothesis must be :-

[tex]H_a:p>0.25[/tex]

Hence, the correct set of hypotheses is

[tex]H_0:p\leq0.25[/tex]

[tex]H_a:p>0.25[/tex]

Answer:

111,952,800 beats in 3 years

Step-by-step explanation:

71 beats/minute, 60 minutes/hour ~ 71x60=4,260 beats/hour

24 hours/day ~ 4,260x24=102,240 beats/day

365 days/year ~ 102,240x365=37,317,600 beats/ year

37,317,600x3=111,952,800 beats in 3 years

The heart beats 111952800 times in 3 years

From the given question, we just have to find the rate at which the heart beats.

Given;

we can start by solving how many minutes are in 1 year.

To do that, we have to multiply 60 minutes by 24 hours by 365 days

[tex]60*24* 365=525600\\ [/tex]

We have 525600 minutes in 1 year

Now, we can multiply this value by 71 to know the number of beats in 1 year.

[tex]525600 * 71 = 37317600[/tex]

The heart beats for 37317600 times in a year.

Let's multiply this value by 3 to know how many times it beats in 3 years.

[tex]37317600 * 3 = 11952800[/tex]

The heart beats 11952800 times in 3 years.

We are also asked to calculate 3.0, 3.00 and 3.000 years

In this case, 3.0 = 3.00 = 3.000  and the rate at which the heart beats is uniform or equal across the three times given.

learn more about rates here;

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Answer:

225

Step-by-step explanation:

To find your answer, multiply 300 by the decimal form of 75%, which is 0.75.

[tex]300 * 0.75 = 225[/tex]

Answer:

Half of 300 is 150 and half of 150 is 75, which is 25%. 75 x 3 gives us 225. 225 is the answer

Step-by-step explanation:

there are 52 cards in a deck, 12 of those cards are "face cards", so the remaining are number cards, namely 40.

52 = sample space

40 = favorable outcomes

P(number card | number card) = p(number) * p(number)

so the first time we pull one, there are 52 cards, the probability of a number card is 40/52, or 10/13, and we don't put it back in the deck.

the next time we pull another card, the cards are no longer 52 total, we pulled one out, they're only 51, namely 51 = sample space, and the number cards if we really pulled out before, are no longer 40, are 39, namely 39 = favorable outcomes.

probability of getting a number card the second time?  39/51 or 13/17.

[tex]\bf \stackrel{\textit{probability of getting a number card twice}}{\cfrac{10}{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{17}\implies \cfrac{10}{17}~~\approx ~~ 0.59}~\hfill 59\%[/tex]

Answer:

105/221

Step-by-step explanation:

There are 52 cards in a deck.

Assuming 2-10 are the number cards

2,3,4,5,6,7,8,9,10 = 9

There are 4 suits

9*4 = 36 cards are number cards

P(1st card is a number card) = number card/ total

                                               =36/52 = 9/13

We do not replace the card, so there are only 51 cards left, and only 35 number cards

P(2nd card is a number card) = number card/ total

                                               =35/51

The probability of getting 2 number cards in a row is

P (number ,number) =P(1st card is a number)*P(2nd card is a number card)  

                                  = 9/13 * 35/51

Dividing the top and bottom by 3      

                                  = 3/13 * 35/17                  

                                  =105/221

Answer:

The Cpk is 0.952

Step-by-step explanation:

The formula to calculate the Cpk of a process is

[tex]Cpk = min(\frac{USL-mean}{3*sigma}, \frac{mean-LSL}{3*sigma} )[/tex]

where

USL (Upper Specification Limit) =3.5cm+0.9cm = 4.4cm

LSL (Lower Specification Limit) =3.5cm-0.9cm=2.6cm

Standard Deviation = sigma = 0.28cm

Mean = 3.4cm

So,

[tex]Cpk=min(\frac{4.4-3.4}{3*0.28} ,\frac{3.4-2.6}{3*0.28})\\\\Cpk=min(\frac{1}{0.84} ,\frac{0.8}{0.84})\\\\Cpk=min(1.190 ,0.952)\\\\\\[/tex]

The Cpk is 0.952

The process capability index or CPk is calculated using the formula min([USL - m]/3σ, [m - LSL]/3σ). In this case, USL is calculated as 4.4 cm and LSL is found to be 2.6 cm. The final CPk value is the smaller of the two resulting values, which in this case is 0.952.

The question asks for the calculation of CPk, which is an index in statistics determining the potential capability of a process in meeting the specification limits. This index considers both the variability of the process and the target in its calculation. The formula for CPk is given by

CPk = min([USL - m]/3σ, [m - LSL]/3σ)

where:

Using the given values from the question,

USL = 3.5 cm + 0.9 cm = 4.4 cm,

LSL = 3.5 cm - 0.9 cm = 2.6 cm,

[USL - m]/3σ = (4.4 cm - 3.4 cm) / (3 * 0.28 cm) = 1.19,

[m - LSL]/3σ = (3.4 cm - 2.6 cm) / (3 * 0.28 cm) = 0.952.

The CPk value will be the smaller of these two values, which is 0.952.

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Final answer:

Calculate the probability that the mean life span of a group of 10 hot water heaters is between 12 and 15 years using the standard normal distribution.

Explanation:

The probability that the mean life span in a group of 10 randomly selected hot water heaters is between 12 and 15 years can be calculated using the standard normal distribution.

Given: Mean = 13 years, Standard Deviation = 1.5 years.

[tex]x=2[/tex]

Represent the sentence mathematically. [tex]2(x-3)=-2[/tex]

Distribute. [tex]2x+(2*-3)=2x-6=-2[/tex]

Add 6 on both sides. [tex]2x=-2+6=4[/tex]

Divide both sides by 2. [tex]x=2[/tex]

Final answer:

To solve the given differential equation y'''+4y''-16y'-64y=0 with initial conditions, we can use the characteristic equation method. By finding the roots of the characteristic equation and applying the initial conditions, the general solution is obtained as y(t) = (-16/21)e^(-8t) + (8/21)e^(2t) + (8/21)e^(-4t).

Explanation:

To solve the given differential equation, we can use the characteristic equation method. We first find the characteristic equation by substituting y = e^(mt) into the differential equation, which gives us the equation (m^3 + 4m^2 - 16m - 64)e^(mt) = 0. Since e^(mt) is never zero, we can simplify the equation to m^3 + 4m^2 - 16m - 64 = 0.

Using a numerical method or factoring, we find that the roots of the characteristic equation are m = -8, m = 2, and m = -4. Therefore, the general solution to the differential equation is y(t) = c1e^(-8t) + c2e^(2t) + c3e^(-4t), where c1, c2, and c3 are constants determined by the initial conditions.

Using the given initial conditions y(0) = 0, y'(0) = 26, and y''(0) = -16, we can solve for the constants. Substituting t = 0 into the general solution and its derivatives, we get the equations c1 + c2 + c3 = 0, -8c1 + 2c2 - 4c3 = 26, and 64c1 + 4c2 + 16c3 = -16. Solving these equations, we find c1 = -16/21, c2 = 8/21, and c3 = 8/21.

Therefore, the solution to the differential equation is y(t) = (-16/21)e^(-8t) + (8/21)e^(2t) + (8/21)e^(-4t).

[tex]f(4+h)-f(4)=(4+h)^2-10-(4^2-10)\\f(4+h)-f(4)=16+8h+h^2-10-16+10\\f(4+h)-f(4)=h^2+8h[/tex]

Answer:

[tex]f (4 + h) -f (4) = h ^ 2 + 8h[/tex]

Step-by-step explanation:

We have the following quadratic function.

[tex]f (x) = x ^ 2-10[/tex]

We must find the following expression

[tex]f (4 + h) -f (4) =[/tex]

First we must find [tex]f (4 + h)[/tex]

Then substitute [tex]x = (4 + h)[/tex] in the quadratic equation:

[tex]f (4 + h) = (4 + h) ^ 2 -10\\\\f (4 + h) = 16 + 8h + h ^ 2 -10\\\\f (4 + h) = h ^ 2 + 8h +6[/tex]

Now we find [tex]f(4)[/tex]. Replace [tex]x = 4[/tex] in the function [tex]f (x)[/tex]

[tex]f (4) = (4) ^ 2-10\\\\f (4) = 16-10\\\\f (4) = 6[/tex]

Finally we have to:

[tex]f (4 + h) -f (4) = h ^ 2 + 8h +6 - 6[/tex]

[tex]f (4 + h) -f (4) = h ^ 2 + 8h[/tex]

Answer:

4 Pairs.

Explanation:

A vertical angle is a set of two opposite angles, they show up when two lines intersect. Their sum is also 180°.

Answer:

4 Pairs

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

Given:

Standard deviation or variance of two population.

We need to write a method by which a formal hypothesis test can be conducted of claim made about two population standard deviations or variances.

In General Chi-Square test and F-test are used for variance or standard deviation.

Also Chi-Square test and F-test require that the original population be normally distributed.

Now for Testing a Claim about Variance or Standard Deviation

To test a claim about the value of the variance or the standard deviation of population, then we use the test statistic which follows chi-square distribution with n − 1 degrees of freedom, and is given by the following formula.

[tex]\chi^2=\frac{(n-1)s^2}{\sigma_0^2}[/tex]

Where s is for given standard deviation and [tex][\sigma[/tex] is for claimed standard deviation.

First we make Hypothesis, then we choose the value of α ( level of significance ), after that using above formula we find value of chi-square.

then we find table value for the chosen α also known as table value or p-value. Finally we give final answer by checking relation between p-value and α.

If  p-value < α then null hypothesis is rejected

If p-value > α  then null hypothesis accepted.

Answer:

[tex]\frac{6x^2+3}{2x^3+3x}[/tex]

Step-by-step explanation:

You need to apply the chain rule here.

There are few other requirements:

You will need to know how to differentiate [tex]\ln(u)[/tex].

You will need to know how to differentiate polynomials as well.

So here are some rules we will be applying:

Assume [tex]u=u(x) \text{ and } v=v(x)[/tex]

[tex]\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}[/tex]

[tex]\text{ power rule } \frac{d}{dx}x^n=nx^{n-1}[/tex]

[tex]\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}[/tex]

[tex]\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}[/tex]

Those appear to be really all we need.

Let's do it:

[tex]\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)[/tex]

[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))[/tex]

[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})[/tex]

[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))[/tex]

[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)[/tex]

[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}[/tex]

I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.

Your answer is [tex]\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}[/tex].

The average wait time in line at Renuka Jain's Car Wash is approximately 7.69 minutes, and the average number of customers waiting in line is approximately 1.54 cars.

Given that Renuka​ Jain's Car Wash takes a constant time of 3.0 minutes for its automated car wash cycle, and cars arrive following a Poisson distribution at the rate of 12 cars per hour (meaning one car every 5 minutes on average), we can calculate the average wait time in line and the average number of customers waiting in line.

To find the average wait time, we use the formula for the wait time in a M/M/1 queue: W = 1/(μ - λ), where λ is the arrival rate and μ is the service rate. We have λ = 12 cars/hour = 0.2 cars/minute, and μ = 1 car/3 mins = 0.33 cars/minute. Thus, the average wait time is W = 1/(0.33 - 0.2) = 7.69 minutes.

For the average number of customers in the line, we use the formula L = λW, where L is the average number of customers in the line, λ is the arrival rate and W is the average wait time. L = 0.2 cars/minute * 7.69 minutes = 1.54 cars.

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Answer:

Dimensions of the container should be 12×12×7.5 ft to minimize the making cost.

Step-by-step explanation:

A trash company is designing an open top, rectangular container having volume = 1080 ft³

Let the length of container = x ft , width of the container = y ft and height of the container = z ft.

So volume of the rectangular container = xyz = 1080 ft³

Or [tex]z=\frac{1080}{xy}[/tex] ft -----(1)

Cost of making the bottom of the container = $5 per square ft

Area of the bottom = xy

Cost of making the bottom @ $5 per square ft = 5xy

Area of all sides of the container = 2(xz + yz) = 2z(x+ y)

Now it has been given that cost of making all sides of the container is = $4 per square ft

So total cost to manufacture sides = 4[2z(x + y)]

Now cost of making bottom and sides of the container = 5xy + 8z(x + y)

We put the value of z from equation 1

Total cost A = 5xy+8(x + y)[tex](\frac{1080}{xy})[/tex]

A = 5xy +[tex]8(\frac{1080}{y})+8(\frac{1080}{x})[/tex]

Now we will find the derivative of A and equate it to the zero

[tex]\frac{dA}{dx}=0[/tex] and [tex]\frac{dA}{dy}=0[/tex]

[tex]\frac{dA}{dx}=5y+8(1080)(0)+8(1080)(-\frac{1}{y^{2}})=0[/tex]

5y =[tex]\frac{8\times1080}{y^{2} }[/tex]

5y³ = 8640

y³ =[tex]\frac{8640}{5}=1728[/tex]

y = 12 ft

For [tex]\frac{dA}{dy}=0[/tex]

[tex]\frac{dA}{dy}=5x+\frac{8(-1080)}{x^{2}}[/tex]=0

5x =[tex]\frac{8(1080)}{x^{2} }[/tex]

5x³ = 8640

x³ = 1728

x = 12

Now from equation 1

z =[tex]\frac{1080}{x}[/tex]

  =[tex]\frac{1080}{144}[/tex]

z = 7.5

Therefore, dimensions of the container should be 12×12×7.5 ft to minimize the making cost.

Final answer:

To minimize the total cost, we need to minimize the cost of the bottom and the cost of the sides. We can find the dimensions of the container that will minimize the total cost by solving a system of equations.

Explanation:

Let's assume that the length of the rectangular container is x ft, the width is y ft, and the height is z ft.

The volume of the container is given as 1080 ft3.

Therefore, we have the equation:

x * y * z = 1080

The cost of making the bottom of the container is $5 per square foot and the cost of the sides is $4 per square foot.

The cost of the bottom is 5 * (x * y).

The cost of the sides is 4 * (2xy + 2xz + 2yz).

To minimize the total cost, we need to minimize the cost of the bottom and the cost of the sides.

First, let's solve the volume equation for x:

x = (1080) / (y * z)

Substituting the value of x in the cost equation, we have:

Cost = 5 * (1080) / (y * z) * y + 4 * (2 * (1080 / (y * z)) * y + 2 * (1080) / (y * z) * z + 2 * y * z)

Now, we can find the minimum cost by taking the derivative of the cost equation with respect to y and z, and setting it equal to zero.

Then, we solve the resulting system of equations to find the values of y and z that minimize the cost.

Finally, we substitute the values of y and z back into the volume equation to find the value of x.

By solving the equations, we can find the dimensions of the container that will minimize the total cost.

Answer: d. 10

Step-by-step explanation:

We know that the number of combinations of r objects selected from a group of n objects at a time is given by :-

[tex]^nC_r=\dfrac{n!}{(n-r)!r!}[/tex]

Given : The total number of letters = 5

The number of letters need to select = 2

Then , the number of combinations of 2 letters selected from a group of 5 letters at a time is given by :-

[tex]^5C_2=\dfrac{5!}{(5-2)!2!}=\dfrac{5\times4\times3!}{3!\times2}=10[/tex]

Hence, there are 10 possible selections.

The problem pertains to combinations in mathematics. When you select two letters out of five without considering the order, you use a formula of 'C(n, r) = n! / [(n-r)!r!]'. Applying this to our problem (where n=5, r=2), it gives us 10 combinations.

The problem you're asking about is associated with combinations in combinatorial mathematics. When selecting two letters out of five (A, B, C, D, and E), we are interested in different combinations and not the order in which you select them. The standard formula to calculate combinations is C(n, r) = n! / [(n-r)!r!].

Here, n = 5 (total number of letters), and r = 2 (the number of letters you want to select). So, C(5, 2) = 5! / [(5-2)!2!] = 10. The correct answer is d. 10.

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Answer:

[tex]\frac{1}{y^2}=2x^3+Cx^4[/tex].

Step-by-step explanation:

Given differential equation

[tex]\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{2}{x}y=x^2y^3[/tex]

Differential equation can be write as

[tex]y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{2}{x}y^{-2}=x^2[/tex]

By Bernoulli method

Susbstitute [tex]y^{-2}=t[/tex].....{equationI}

Differentiate equation I w.r.t x then we get

[tex]\frac{\mathrm{d}t}{\mathrm{d}x}=-2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}[/tex]

[tex]-\frac{1}{2}\frac{\mathrm{d}t}{\mathrm{d}x}=y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}[/tex]

Susbstitute the values in the given differential equation then we get

[tex]-\frac{1}{2}\frac{\mathrm{d}t}{\mathrm{d}x}+\frac{2}{x}t=x^2[/tex]

[tex]\frac{\mathrm{d}t}{\mathrm{d}x}-\frac{4}{x}t=-2x^2[/tex]

It is first order linear differential equation and compare with the first order linear differential equation [tex]\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)[/tex]

Then we get P(x)=[tex]-\frac{4}{x}[/tex] and Q(x)=[tex]-2x^2[/tex]

Integration factor=[tex]e^\intP(x)dx[/tex]

Integration factor= [tex]e^{-\int\frac{4}{x}dx[/tex]

Integration factor= [tex]e^{-4lnx}=e^{lnx^{-4}}=x^{-4}[/tex].

Using [tex]e^{logb}=b[/tex]

[tex]t\times \frac{1}{x^4}=\int{-2x^2}\times\frac{1}{x^4}dx+C[/tex]

[tex]t=-2x^4{\intx^{-2}dx+C}[/tex]

[tex]t=2x^4\times\frac{1}{x}+Cx^4[/tex]

[tex]t=2x^3+Cx^4[/tex]

Substitute [tex]t=\frac{1}{y^2}[/tex] then we get

[tex]\frac{1}{y^2}=2x^3+Cx^4[/tex].

Answer: [tex]\frac{1}{y^2}=2x^3+Cx^4[/tex].

Answer:

cross sectional area of the wing's is = 3404.8 cm²

Step-by-step explanation:

using n= 5 to estimate area of the wing's

a = 160

taking sum of thickness at n = 1, 3, 5, 7, 9

so sum of the measurement of the thickness at the given position

19.9 +29.0 + 27.5 +20.9 + 9.1 = 106.4

so the thickness is 106.4/5

           = 21.28 cm

cross sectional area of the wing's is = 160 × 21.28

                                                            = 3404.8 cm²

Using the Midpoint Rule with 5 intervals, the estimated area of the airplane wing's cross-section can be obtained by dividing the total span into equal parts, calculating the midpoints of the measurements, and then adding up these individual areas.

To answer this question, we need to apply the Midpoint Rule - a method used in mathematics for approximating the definite integral of a function. The Rule works by estimating the area under the curve by rectangles, whose heights are determined by the function values at the midpoints of their bases.

Given n = 5, we divide the total measurement span (160 cm) into 5 parts. So, each part/subinterval is 32 cm.

We calculate the area of each part by multiplying its width (32 cm) by its midpoint height. For a sequence of measurements, the midpoints are obtained by averaging two consecutive measurements.

The midpoints for the given measurements are:

We then sum up the areas of all parts to get the estimated area of the airplane wing's cross-section.

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Answer:

10. México es un país con mucha gente (más de 130 millones de personas). Aunque es más grande, la Argentina tiene menos 45 millones de personas.

ARGENTINA NO ES TAN GRANDE COMO MÉXICO.

11. Ellos tienen cinco hijos y nosotros otros tenemos cinco hijos también.

ELLOS TIENEN TANTOS HIJOS COMO NOSOTROS.

12. Carlos no tiene mucho dinero, pero Felipe es rico.

CARLOS NO TIENE TANTO DINERO COMO FELIPE.

13. Linda es muy simpática, me gusta Dolores.

LINDA ES TAN SIMPÁTICA, PERO ME GUSTA DOLORES.

Answer:

1120 possible ways

Step-by-step explanation:

In order to find the answer we need to be sure what equation we need to use.

From the given example, let's consider initially only men. Because you have a total of 8 men and we need to chose only 3 men, let's suppose that the 3 chosen men are A, B, and C.

Because A,B,C is the same as choosing C,B,A, which means it doesn't matter the order of the chosen men, we need to use a 'combination equation'.

Because we have two groups (women and men) then we have:

Possible ways = 8C3 * 6C3 (which are the combinations for women and men respectively). Remember that:

nCk=n!/((n-k)!*k!) so:

Possible ways = 8!/((8-3)!*3!) * 6!/((6-3)!*3!) = 56* 20 = 1120.

In conclusion, there are 1120 possible ways.

Answer:

3x - y = 0; 2x - y = -5

Step-by-step explanation:

Let x be the present age of Hans and y be the present age of Grace,

Since, in present Grace is three times as old as Hans,

⇒ y = 3x

⇒ 3x - y = 0

Now, after 5 years,

The age of Hans = x + 5,

And, the age of Grace = y + 5

Also, in 5 years Grace will be twice as old as Hans is then,

⇒ y + 5 = 2 ( x + 5 )

⇒ y + 5 = 2x + 10

⇒ 2x - y = -5

Hence, the required system of linear equations is,

3x - y = 0; 2x - y = -5

Answer:

D.368 space ft squared

Step-by-step explanation:

Hello

The equation to find the area of ​​the rectangle is simply A = h * b. This means that the area of ​​a rectangle is equal to the product of its height (h) by its base (b), or of its length by its width

Let

A=h*b

h=23

b=16

A=23 ft*16 ft

A=368 ft squared

so, the answer is

D. 368 ft squared.

I hope it helps

Have a fantastic day.