Buses headed to Longmont arrive in downtown Boulder every 30 minutes starting at 8:37am, whereas buses heading to Denver arrive 15 minutes starting at 8:31am. (a) If a passenger arrives at the station uniformly between 8:30am and 9:30am and then gets onto the first bus that arrives, what is the probability she goes to Longmont? (b) What is the passenger arrives uniformly between 8:45am and 9:45am?

Answer with explanation:

Let m and n are integers

To prove that m+n and m-n are either both even or both odd.

1. Let m and n are both even

We know that sum of even number is even and difference of even number is even.

Suppose m=4 and n=2

m+n=4+2=6 =Even number

m-n=4-2=2=Even number

Hence, we can say m+n and m-n are both even .

2. Let m and n are odd numbers .

We know that sum of odd numbers is even and difference of odd numbers is even.

Suppose m=7 and n=5

m+n=7+5=12=Even number

m-n=7-5=2=Even number

Hence, m+n and m-n are both even .

3. Let m is odd and n is even.

We know that sum of an odd number and an even number is odd and difference of an odd and an even number is an odd number.

Suppose m=7 , n=4

m+n=7+4=11=Odd number

m-n=7-4=3=Odd number

Hence, m+n and m-n are both odd numbers.

4.Let m is even number and n is odd number .

Suppose m=6, n=3

m+n=6+3=9=Odd number

m-n=6-3=3=Odd number

Hence, m+n and m-n are both odd numbers.

Therefore, we can say for all inetegers m and n , m+n and m-n are either both even or both odd.Hence proved.

Final answer:

The problem is solved by expressing the conditions under which m and n are both even or odd, and their sum and difference in terms of 2k (for even) and 2k+1 (for odd), demonstrating that m+n and m-n are both even or both odd.

Explanation:

To prove that for all integers m and n, m + n and m - n are either both even or both odd, we start by recalling the definition of even and odd numbers. An even number can be expressed as 2k, where k is an integer, and an odd number can be expressed as 2k + 1, where k is an integer.

This argument shows that m + n and m - n must either be both even or both odd for any integers m and n.

Answer: Null hypothesis = [tex]H_0:p=0.74[/tex]

Alternative hypothesis = [tex]H_1:p\neq0.24[/tex]

Step-by-step explanation:

Given claim : 74% of workers got their job through college.

In proportion , 0.74 of workers got their job through college.

Let p be the proportion of workers got their job through college.

Then claim : [tex]p=0.74[/tex]

We know that the null hypothesis always takes equality sign and alternative hypothesis takes just opposite of the null hypothesis.

Thus, Null hypothesis = [tex]H_0:p=0.74[/tex]

Alternative hypothesis = [tex]H_1:p\neq0.24[/tex]

Answer:

$1210

Step-by-step explanation:

Let x be total amount

First John spent $110 on a radio and 4/11 of what was left on presents for his friends so he was left with

[tex]\frac{7}{11}(x-110)=\frac{7}{11}x-70[/tex]

Then he put 2/5 of his remaining money into a checking account

[tex]\frac{3}{5}\left(\frac{7}{11}x-70\right)[/tex]

Rest he donated to charity

[tex]420=\frac{3}{5}\left(\frac{7}{11}x-70\right)\\\Rightarrow \left(\frac{7}{11}x-70\right)=\frac{2100}{3}\\\Rightarrow \frac{7}{11}x-70=700\\\Rightarrow \frac{7}{11}x=770\\\Rightarrow x=1210[/tex]

Hence total amount of money John originally had was $1210

Answer:

a) 0.7898

b) 0.0516

c) 0.1587

Step-by-step explanation:

Given : Mean : [tex]\mu=8.1\text{ minutes}[/tex]

Standard deviation : [tex]\sigma =1.9\text{ minutes}[/tex]

Since , the police response time has a normal distribution.

The formula to calculate the z-score :-

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

For x=5 minutes.

[tex]z=\dfrac{5-8.1}{1.9}=-1.63[/tex]

For x=10 minutes.

[tex]z=\dfrac{10-8.1}{1.9}=1[/tex]

a) The p-value =[tex]P(-1.63<z<1)=P(z<1)-P(z<-1.63)[/tex]

[tex]=0.8413447-0.0515507=0.789794\approx0.7898[/tex]

b) The p-value =[tex]P(z<-1.63)=0.0515507\approx0.0516[/tex]

c) The p-value =[tex]P(z>1)=1-P(z<1)[/tex]

[tex]=1-0.8413447=0.1586553\approx0.1587[/tex]

We calculated the probability of different police response times using the z-score method. The probability of a response time between 5 and 10 minutes is 0.7897, the probability for less than 5 minutes is 0.0516, and more than 10 minutes is 0.1587.

To answer this question, we need to first standardize the response times using the z-score formula: z = (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation.

(a) To find the probability that the response time is between 5 and 10 minutes, we first calculate the z-scores for 5 and 10 minutes:

Z(5) = (5 - 8.1) / 1.9 = -1.632 Z(10) = (10 - 8.1) / 1.9 = 1

Next, we find these values in the z-table which yields: P(Z<1) = 0.8413, P(Z<-1.632) = 0.0516. The probability that the response time is between 5 and 10 minutes is the difference between these values, so P(5 < X < 10) = 0.8413 - 0.0516 = 0.7897.

(b) For the response time less than 5 minutes, we calculate the probability using the z-score for 5 minutes. Z(5) = -1.632, looking in the z-table, we find this value equals to 0.0516. Therefore, the response time is less than 5 minutes is 0.0516.

(c) Lastly, the probability for a response time more than 10 minutes is P(Z > 1) which is equal to 1 - P(Z < 1). From the z-table, we find P(Z<1) = 0.8413. Then, P(Z > 1) = 1 - 0.8413 = 0.1587. So, the probability that the response time is more than 10 minutes is 0.1587.

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Answer:[tex]\frac{22}{3}[/tex],[tex]2\dot \sqrt{\frac{11}{3}}[/tex]

Step-by-step explanation:

Given

rectangle with its base on x-axis

and other two corners at parabola

and parabola is downward facing symmetric about y-axis

let y be the y co-ordinate of the corner thus x co-ordinate is given by

[tex]x=\pm \sqrt{11-y}[/tex]

Thus lengths of rectangle is [tex]2\sqrt{11-y}[/tex] & y

Area [tex]=y\times 2\sqrt{11-y}[/tex]

differentiating w.r.t to y for maximum area

[tex]\frac{\mathrm{d} A}{\mathrm{d} y}=2\times \sqrt{11-y}-\frac{y}{2\dot \sqrt{11-y}}=0[/tex]

we get y=[tex]\frac{22}{3}[/tex]

and [tex]x=\pm \sqrt{\frac{11}{3}}[/tex]

A_{max}=16.21 units

Answer:

So anyways the equation appears to be [tex]y=(x+4)^3+3[/tex].

Step-by-step explanation:

It looks like a cubic to me.

That is the parent function looks like [tex]f(x)=x^3[/tex].

I'm going to identify the transformations here by using the zero of the function I called f.  So where has the point (0,0) on [tex]y=x^3[/tex] wonder to in the new graph.  It appears to be (-4,3).  So the graph moved left 4 units and up 3 units.

f(x+4)+3 moves the graph left 4 and up 3.

----------Other notes:

f(x-4)+3 moves the graph right 4 and up 3.

f(x+4)-3 moves the graph left 4 and down 3.

f(x-4)-3 moves the graph right 4 and down 3.

So anyways the equation appears to be [tex]y=(x+4)^3+3[/tex].

To determine the equation of a transformed basic function, we typically look at how the graph's shape compares to one of the standard basic functions. The basic functions include linear functions, quadratic functions, absolute value functions, square root functions, cubic functions, and exponential and logarithmic functions. We also consider basic trigonometry functions for periodic graphs.
To find the equation, we need to identify four main transformations that may have been applied to the basic function:
1. **Vertical stretching/shrinking**: If the graph is stretched or shrunk vertically, this is represented by a multiplication factor `a` in front of the basic function `f(x)`.
2. **Horizontal stretching/shrinking**: If the graph is stretched or shrunk horizontally, this is represented by a factor within the function's argument, such as `f(bx)`, where `1/b` is the stretching/shrinking factor.
3. **Vertical shifting**: If the graph is shifted up or down, a constant `c` is added or subtracted from the function, giving `f(x) + c`.
4. **Horizontal shifting**: If the graph is shifted left or right, the function's input is adjusted by adding or subtracting a constant `d` within the argument of the function, yielding `f(x - d)`.
5. **Reflections**: If the graph is flipped over the x-axis, this is represented by a negative sign in front of the `a` factor. If it's flipped over the y-axis, the negative sign is inside the function's argument, `f(-x)`.
Without any specific details about the graph's appearance, the points, or what basic function it resembles, it is impossible to provide the exact transformed function. However, for illustrative purposes, I’ll demonstrate how one might find the equation for a transformed quadratic function based on hypothetical graph observations:
Suppose you find that the graph looks like a parabola that opens upwards and has been:
- Stretched vertically by a factor of 3 (vertical stretch)
- Compressed horizontally by a factor of 1/2 (horizontal stretch)
- Shifted up by 5 units (vertical shift)
- Shifted to the right by 4 units (horizontal shift)
With these observations, you would start with the standard quadratic function, `f(x) = x^2`, and apply the transformations:
1. Vertical stretch by 3: `f(x) = 3x^2`
2. Horizontal compression by a factor of 1/2, which is equivalent to stretching by a factor of 2: `f(x) = 3(x/2)^2 = 3(x^2/4) = (3/4)x^2`
3. Vertical shift up by 5 units: `f(x) = (3/4)x^2 + 5`
4. Horizontal shift right by 4 units: `f(x) = (3/4)(x - 4)^2 + 5`
The transformed function based on the hypothetical scenario would be `f(x) = (3/4)(x - 4)^2 + 5`.
Without specifics of the graph in question, you would follow a similar process: identify the basic function type based on the shape of the graph and apply the relevant transformations.

a. The marginal densities

[tex]f_X(x)=\displaystyle\int_0^1(x+y)\,\mathrm dy=x+\frac12[/tex]

and

[tex]f_Y(y)=\displaystyle\int_0^1(x+y)\,\mathrm dx=y+\frac12[/tex]

b. This can be obtained by integrating the joint density over [0.25, 1] x [0.5, 1]:

[tex]P(X>0.25,Y>0.5)=\displaystyle\int_{1/4}^1\int_{1/2}^1(x+y)\,\mathrm dx\,\mathrm dy=\frac{33}{64}[/tex]

Final answer:

To find the marginal distributions of X and Y, we integrate the joint density function over the range of the other variable. The marginal distribution of X is f(x) = x+1/2, for 0≤x≤1. The marginal distribution of Y is f(y) = y+1/2, for 0≤y≤1.

Explanation:

To find the marginal distributions of X and Y, we need to integrate the joint density function over the range of the other variable. For the marginal distribution of X, we integrate f(x,y) with respect to y from 0 to 1:

(∫⁰₁(x+y) dy) = (x+y/2)∣⁰₁ = x+1/2

So, the marginal distribution of X is given by f(x) = x+1/2, for 0≤x≤1.

Similarly, for the marginal distribution of Y, we integrate f(x,y) with respect to x from 0 to 1:

(∫⁰₁(x+y) dx) = (x2/2+xy)∣⁰₁ = y+1/2

Therefore, the marginal distribution of Y is given by f(y) = y+1/2, for 0≤y≤1.

Apparently this solution is too long for posting, so I've written it elsewhere and am attaching screenshots of it.

The first four terms of each solution are

[tex]1-\dfrac23x^3+\dfrac5{36}x^6-\dfrac{10}{567}x^9[/tex]

and

[tex]x-\dfrac1{24}x^4+\dfrac1{315}x^7-\dfrac7{32,400}x^{10}[/tex]

Answer:

Total number of ways will be 209

Step-by-step explanation:

There are 6 boys and 4 girls in a group and 4 children are to be selected.

We have to find the number of ways that 4 children can be selected if at least one boy must be in the group of 4.

So the groups can be arranged as

(1 Boy + 3 girls), (2 Boy + 2 girls), (3 Boys + 1 girl), (4 boys)

Now we will find the combinations in which these arrangements can be done.

1 Boy and 3 girls = [tex]^{6}C_{1}\times^{4}C_{3}=6\times4[/tex]=24

2 Boy and 2 girls=[tex]^{6}C_{2}\times^{4}C_{2}=\frac{6!}{4!\times2!}\times\frac{4!}{2!\times2!}=15\times6=90[/tex]

3 Boys and 1 girl = [tex]^{6}C_{3}\times^{4}C_{1}=\frac{6!}{4!\times2!}\times\frac{4!}{3!}=\frac{6\times5\times4}{3 \times2} \times4=80[/tex]

4 Boys = [tex]^{6}C_{4}=\frac{6!}{4!\times2!} =\frac{6\times 5}{2\times1}=15[/tex]

Now total number of ways = 24 + 90 + 80 + 15 = 209

Answer with explanation:

The given statement is which we have to prove by the principal of Mathematical Induction

    [tex]2^{n}>n[/tex]

1.→For, n=1

L H S =2

R H S=1

2>1

L H S> R H S

So,the Statement is true for , n=1.

2.⇒Let the statement is true for, n=k.

      [tex]2^{k}>k[/tex]

                   ---------------------------------------(1)

3⇒Now, we will prove that the mathematical statement  is true for, n=k+1.

     [tex]\rightarrow 2^{k+1}>k+1\\\\L H S=\rightarrow 2^{k+1}=2^{k}\times 2\\\\\text{Using 1}\\\\2^{k}>k\\\\\text{Multiplying both sides by 2}\\\\2^{k+1}>2k\\\\As, 2 k=k+k,\text{Which will be always greater than }k+1.\\\\\rightarrow 2 k>k+1\\\\\rightarrow2^{k+1}>k+1[/tex]

Hence it is true for, n=k+1.

So,we have proved the statement with the help of mathematical Induction, which is

      [tex]2^{k}>k[/tex]

Answer:

130 cars.

Step-by-step explanation:

The cost function is given by:

C(x) = 0.9x^2 -234x + 23,194; where x is the input and C is the total cost of production.

To find the minimum the unit cost, there must be a certain number of cars which have to be produced. To find that, take the first derivative of C(x) with respect to x:

C'(x) = 2(0.9x) - 234 = 1.8x - 234.

To minimize the cost, put C'(x) = 0. Therefore:

1.8x - 234 = 0.

Solving for x gives:

1.8x = 234.

x = 234/1.8.

x = 130 units of cars.

To check whether the number of cars are minimum, the second derivative of C(x) with respect to x:

C''(x) = 1.8. Since 1.8 > 0, this shows that x = 130 is the minimum value.

Therefore, the cars to be made to minimize the unit cost = 130 cars!!!

Answer:

1. { a, b, c, d, e, h }

2. { f, g, i }

Step-by-step explanation:

Given sets,

A = {a, b, c, d, e},

B = {a, c, f, g, i}

Universal set , ∪ = {a, b, c, d, e, f, g, h, i},

1. Since, [tex]B^c[/tex] = elements of universal set which are not in set B

=  U - B

= { b, d, e, h },

Thus,

[tex]A\cup B^c[/tex] = All elements of A and [tex]B^c[/tex]

= { a, b, c, d, e, h }

2. B - A = elements of set B which are not in set A

= { f, g, i }

If (2, -1, 2, 1) is a linear combination of the three given vectors, then there should exist [tex]c_1,c_2,c_3[/tex] such that

[tex](2,-1,2,1)=c_1(1,-2,0,3)+c_2(2,3,0,-1)+c_3(3,9,-4,-2)[/tex]

or equivalently, there should exist a solution to the system

[tex]\begin{cases}c_1+2c_2+3c_3=2\\-2c_1+3c_2+9c_3=-1\\-4c_3=2\\3c_1-c_2-2c_3=1\end{cases}[/tex]

Right away we get [tex]c_3=-\dfrac12[/tex], so the system reduces to

[tex]\begin{cases}c_1+2c_2=\dfrac72\\\\-2c_1+3c_2=\dfrac72\\\\3c_1-c_2=0\end{cases}[/tex]

Notice that the first equation is the sum of the latter two. The third equation gives us

[tex]3c_1-c_2=0\implies 3c_1=c_2[/tex]

so that in the second equation,

[tex]-2c_1+3c_2=\dfrac72\implies7c_1=\dfrac72\implies c_1=\dfrac12[/tex]

which in turn gives

[tex]3c_1=c_2\implies c_2=\dfrac32[/tex]

and hence the (2, -1, 2, 1) is a linear combination of the given vectors, with

[tex]\boxed{(2,-1,2,1)=\dfrac12(1,-2,0,3)+\dfrac32(2,3,0,-1)-\dfrac12(3,9,-4,-2)}[/tex]

Answer:

total profit=$607.278

Step-by-step explanation:

company's profit in 1985= $535 million

company's profit in 1990=$570 million

growth rate = [tex]\frac{570-535}{535}\times 100[/tex]

                    = [tex]\frac{35}{535} \times 100[/tex]

                    = 6.54 %

profit in year 1995 will be = [tex]\frac{6.54}{100}\times 570 =\ \$37.278[/tex]

hence total profit= $570+$37.278

                             = $607.278

Answer:

8710 units

Step-by-step explanation:

Step 1: Write all the data

Fixed cost: $9000

Average variable cost: 9.3 per unit

Total cost: 90,000

Total units: x

Step 2: Find the total variable cost

Average variable cost is per unit so it has to be multiplied by the number of units to find the total variable cost.

Total variable cost = average variable cost per unit x number of units

Total variable cost = 9.3x

Step 3: Make the formula for finding x

Total cost = total fixed cost + total variable cost

90,000 = 9000 + 9.3x

81000 = 9.3x

x = 8709.67

Rounded off to 8710 units

!!

Answer:

The 95% confidence interval for the true mean speed of thunderstorms is [10.712, 13.688].

Step-by-step explanation:

Given information:

Sample size = 10

Sample mean = 12.2 mph

Standard deviation = 2.4

Confidence interval = 95%

At confidence interval 95% then z-score is 1.96.

The 95% confidence interval for the true mean speed of thunderstorms is

[tex]CI=\overline{x}\pm z*\frac{s}{\sqrt{n}}[/tex]

Where, [tex]\overline{x}[/tex] is sample mean, z* is z score at 95% confidence interval, s is standard deviation of sample and n is sample size.

[tex]CI=12.2\pm 1.96\frac{2.4}{\sqrt{10}}[/tex]

[tex]CI=12.2\pm 1.487535[/tex]

[tex]CI=12.2\pm 1.488[/tex]

[tex]CI=[12.2-1.488, 12.2+1.488][/tex]

[tex]CI=[10.712, 13.688][/tex]

Therefore the 95% confidence interval for the true mean speed of thunderstorms is [10.712, 13.688].

Answer:

163

Step-by-step explanation:

So n=391.

This means p=23 and q=17 where p*q=n.

[tex] \lambda (391)=lcm(23-1,17-1)=lcm(22,16)=2*8*11=16*11=176. [/tex]

We want to choose e so that e is between 1 and 176 and the gcd(e,176)=1.

There is only one number in your list that is between 1 and 176... Hopefully the gcd(163,176)=1.

It does. See notes below for checking it:

176=2(88)=2(4*22)=2(2)(2)(2)(11)

None of the prime factors of 176 divide 163 so we are good.

The answer is 163.

Answer:

The full budget has a total value of $160,000

Step-by-step explanation:

This is a simple ratio problem and can be solved using the Rule of Three property. The Rule of Three is used to compare two ratios and find the missing part. In this case the ratios would be the following.

[tex]\frac{64,000}{40 percent} = \frac{x}{100 percent}[/tex]

[tex]\frac{64,000 * 100 percent}{40 percent} = \frac{x}[/tex]

[tex]\frac{6,400,000}{40 percent} = \frac{x}[/tex]

[tex]160,000= \frac{x}[/tex]

So now we know that 100% (The full budget) has a total value of $160,000.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:

The probability is 0.995 ( approx ).

Step-by-step explanation:

Let X represents the event of baby girl,

The probability of a baby being a girl is, p = 0.469,

So, the probability of a baby who is not a girl is, q = 1 - 0.469 = 0.531,

Also, the total number of experiment, n = 7

Thus, by the binomial distribution formula,

[tex]P(x)=^nC_x(p)^x q^{n-x}[/tex]

Where, [tex]^nC_x=\frac{n!}{x!(n-x)!}[/tex]

The probability that all babies are girl or there is no baby boy,

[tex]P(X=7)=^7C_7(0.469)^7(0.531)^{7-7}[/tex]

[tex]=0.00499125661758[/tex]

Hence, the probability that at least one of them is a boy​ = 1 - P(X=7)

= 1 - 0.00499125661758

= 0.995008743382

≈ 0.995

Answer:

These two samples are independent samples.

Step-by-step explanation:

Each object would then have two weight values (one from scale 1 and one from scale 2). Based on the nature of the differences in the two weight measurements for the 30 objects, the two scales may be compared.

These two samples are independent samples.

Though the objects are same but the scales are different.

Answer:

The amount after 8 years is $19249.17

Step-by-step explanation:

For any calculation for investments there si the compound interest formula:

[tex]A=P(1+\frac{r}{n} )^(n*t)[/tex]

Where

P = principal amount (the initial amount you borrow or deposit)

r  = annual rate of interest (as a decimal)

t  = number of years the amount is deposited or borrowed for.

A = amount of money accumulated after n years, including interest.

n  =  number of times the interest is compounded per year  

So for this example

P, the original amount ($14000)

r, 4%

t, 8 years

A, the amount after 8 years

n, 4, due that is quarterly

[tex]P=$14000(1+((4/100)/(4)))^(4*8)\\\\P= $19249.17[/tex]

To cover a 6,000 square foot lawn, 192 ounces of fertilizer will be required. With the given cost of the fertilizer, the total cost to fertilize the lawn would be $9.36.

(a) To calculate how many ounces are required to cover a 6,000 square foot lawn, we can set up a proportion.

40 ounces of Lazy Lawn fertilizer covers 1,250 square feet, so let's represent it as 40/1250. We know that we have a 6,000 square foot lawn but we don't know how many ounces it requires, let's represent it as x/6000.

Our proportion will then be as follows: 40/1250 = x/6000. Cross multiply to find 'x', we will get x = [(40*6000)/1250] = 192 ounces.

(b) Now, to calculate the total cost we first need to know how many 24 ounce bags we will need. 192 ounces divided by 24 ounces per bag gives us 8 bags. Then we calculate the cost, 8 bags times $1.17 per bag gives us a total of $9.36.

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Answer with explanation:

Here, a and b are two real numbers.

Arithmetic Mean of a and b

          [tex]A=\frac{a+b}{2}[/tex]

[tex]\rightarrow a< \frac{a+b}{2}<b[/tex]

Geometric Mean of a and b

         [tex]G=\sqrt{ab}[/tex]

[tex]\rightarrow a< \sqrt{ab}<b[/tex]

[tex]A-G\\\\=\frac{a+b}{2}-\sqrt{ab}\\\\=\frac{a+b-2\sqrt{ab}}{2}\\\\=[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}]^2>0\\\\A-G>0\\\\A>G[/tex]

Square of difference of any two numbers is greater than or equal to 0.

⇒A.M of two Numbers > G.M of two Numbers

Answer:

hence rate interest r = 41.176%

Step-by-step explanation:

The amount borrowed= $680

amount payed back = $750

therefore, interest incurred = 750-680= $70

time, t= 3 months = 3/12= 0.25 years

rate%, r

we know that SI= [tex]\frac{PRT}{100}[/tex]

70= [tex]\frac{680\timesr\0.25}{100}[/tex]

r=[tex]\frac{7000}{0.25\times680} = 41.176[/tex]

hence rate interest r = 41.176%

Answer:

Explained

Step-by-step explanation:

Simulation is nothing but an approximate or somewhat accurate imitation of a real world situation. Simulation is actually a computer generated graphics to predict how a system will behave under given set of parameters, without actually applying real resources. Simulation finds a variety of application in various fields.  Simulation of blood flowing through veins and arteries. Simulation of LBW decisions in a cricket match, which helps Umpires to make correct LBW decisions in a match. A lot of recondite  process can understood using Simulation videos that is why concept of smart learning as been introduced.

We have

[tex]\dfrac{\partial(y\cos(xy)+3x^2)}{\partial y}=\cos(xy)-xy\sin(xy)[/tex]

[tex]\dfrac{\partial(x\cos(xy)+2y)}{\partial x}=\cos(xy)-xy\sin(xy)[/tex]

so the ODE is indeed exact. Then there's a solution of the form [tex]f(x,y)=C[/tex] such that

[tex]\dfrac{\partial f}{\partial x}=y\cos(xy)+3x^2[/tex]

[tex]\implies f(x,y)=\sin(xy)+x^3+g(y)[/tex]

Differentiating wrt [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=x\cos(xy)+2y=x\cos(xy)+g'(y)[/tex]

[tex]\implies g'(y)=2y\implies g(y)=y^2+C[/tex]

Then the solution to the ODE is

[tex]f(x,y)=\boxed{\sin(xy)+x^3+y^2=C}[/tex]

Answer:

  6.9%

Step-by-step explanation:

Interest rate is the one variable in an amortization formula that cannot be determined explicitly. An iterative solution is required, which means the computation must be done by a calculator, spreadsheet, or web site.

My TI-84 TVM Solver tells me that for the given loan amount and payment schedule, the APR is about 6.9%.

Answer:

21 % below what she needs

Step-by-step explanation:

She had hit 12, 10 and 9

This averages to

(12+10+9)/3 = 31/3 =10 1/3

She needs 13

Percentage = (needed-actual)/needed * 100%

                    = (13-10 1/3) / 13 * 100%

                    = (2 2/3) /13 * 100%

                    =.205128205 * 100%

                   =20.5128205%

To the nearest whole number

                      21 %

She is 21 % below what she needs

Samantha's current average free throw percentage is 15 percentage points away from the 65% needed to earn her wellness challenge chip.

Samantha is working on improving her free throw percentage in basketball, and she needs to calculate how far her average is from the required target to earn her wellness challenge chip. To determine her average free throw percentage, we first calculate her average number of successful shots by adding her last three attempts and dividing by three: (12 + 10 + 9) / 3 = 31 / 3 = 10.33, rounded to 10 successful shots on average. Since she takes 20 shots each time, her average percentage is (10 / 20) * 100 = 50%.

To earn her chip, she needs to hit 13 out of 20 free throws, which is (13 / 20) * 100 = 65%. The difference between her current average and the needed percentage is 65% - 50% = 15%. Rounding to the nearest whole number, she is 15 percentage points away from the required free throw percentage to succeed in the challenge.

We are asked to prove the statement,

For all sets A, B, and C, if A ⊆ B then A∪C ⊆ B∪C

Let us consider set A as:

         A={1,3,4,5}

and B={1,2,3,4,5,6,7}

Clearly we may observe that A is a subset of B.

( Since, all the elements of set A are contained in set B .

Hence, A is a subset of B)

Now let us consider set C as:

C={1,2}

Hence,

A∪C={1,2,3,4,5}

and

B∪C={1,2,3,4,5,6,7}

Still we observe that:

          A∪C ⊆ B∪C

Since all the elements of the set A∪C are contained in the set B∪C.

m - my age

s - son's age

[tex]s-6=\dfrac{m-6}{3}\\s+6=\dfrac{m+6}{2}\\\\3s-18=m-6\\2s+12=m+6\\\\m=3s-12\\m=2s+6\\\\3s-12=2s+6\\s=18[/tex]

He's 18