Which statement is true regarding the graphed functions?

Answer:

Secretariat's average speed in the Kentucky Derby of 1973 was of 47.05 miles per hour.

Step-by-step explanation:

Secretariat set the current record of 1 minute 59.40 seconds for the Kentucky Derby, running 10 furlongs, which are 1.25 miles.

To determine the speed at which the horse ran, we must first isolate the time in which he ran, to determine the fraction of time in which he traveled the 1.25 mile.

An hour has 60 minutes. To determine the speed, whe have to divide 60 by the minutes the horse ran, and then multiply it by the number of miles he ran: 60 / 1.5940 = 37,64 x 1.25 = 47,05.

The horse ran at 47.05 miles per hour.

Secretariat's record average speed during the Kentucky Derby was an impressive 37.68 mi/hr.

The question asks to calculate Secretariat's average speed during the Kentucky Derby in miles per hour (mi/hr). First, we need to express the distance of the Kentucky Derby in miles. The Derby is 10 furlongs long, and since 1 furlong is 1/8 mile, this translates to 1.25 miles (10 furlongs × 1/8 mile/furlong). Understanding this conversion is crucial because speed is typically measured in miles per hour in the United States.

Next, Secretariat's record-setting time for completing this distance is 1 minute and 59.40 seconds, which is equivalent to 119.40 seconds. To find the average speed, we use the formula:

Speed = Distance / Time

Hence, Secretariat's average speed is:

1.25 miles / (119.40 seconds × (1 hour / 3600 seconds)) = 1.25 miles / 0.03317 hours = 37.68 mi/hr, when the time is converted from seconds to hours.

Therefore, Secretariat's record average speed during the Kentucky Derby was an impressive 37.68 mi/hr.

Answer:   The required solution of the given IVP is

[tex]x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]

Step-by-step explanation:  We are given to use Laplace transforms to solve the following initial value problem :

[tex]x^{\prime\prime}+9x=1,~~~x(0)=0=x^\prime(0).[/tex]

We will be using the following formula for the Laplace transform :

[tex](i)~L\{t^n\}=\dfrac{n!}{s^{n+1}},\\\\\\(ii)~L\{\coskt\}=\dfrac{s}{s^2+k^2}.[/tex]

Applying Laplace transform on both sides of the above equation, we have

[tex]L\{x^{\prime\prime}+9x\}=L\{1\}\\\\\\\Rightarrow s^2X(s)-sx(0)-x^\prime(0)+9X(s)=\dfrac{1}{s}\\\\\\\Rightarrow s^2X(s)-s\times0-0=\dfrac{1}{s}\\\\\\\Rightarrow (s^2+9)X(s)=\dfrac{1}{s}\\\\\\\Rightarrow X(s)=\dfrac{1}{s(s^2+9)}\\\\\\\Rightarrow X(s)=\dfrac{1}{9s}-\dfrac{s}{9(s^2+9)}.[/tex]

Taking inverse Laplace transform on both sides of the above equation, we get

[tex]L^{-1}\{X(s)\}=L^{-1}\left(\dfrac{1}{9s}\right)-L^{-1}\left(\dfrac{s}{9(s^2+9)}\right)\\\\\\\Rightarrow x(t)=\dfrac{1}{9}\times1-\dfrac{1}{9}\times \cos3t\\\\\\\Rightarrow x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]

Thus, the required solution of the given IVP is

[tex]x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]

To solve the given initial value problem using Laplace transforms, apply the transforms to the equation, simplify to get the equation in the s-domain, and then take the inverse Laplace transform to get the solution in the time domain.

To solve the initial value problem x" + 9x = 1 where x(0) = 0 = x'(0) using Laplace transforms, we firstly apply it to the whole equation. The Laplace transform of a second derivative x''(t) is s^2 X(s) - s x(0) - x`(0) and that of x(t) is X(s), where X(s) is the Laplace transform Of x(t). Since the Laplace transform of a constant is the constant itself divided by s, the Laplace transform of the equation becomes s^2 X(s) - 0 - 0 + 9X(s) = 1/s. Simplifying the above gives X(s) = 1/(s^3 + 9s), as a solution in the s-domain. Apply a combination of basic formulas and decomposition techniques to convert this equation into a zero-state solution. After this, take the inverse Laplace transform to get the solution in the time domain. It should be simple given that the initial values are zero.

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[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(5-1)^2+(1-4)^2}\implies d=\sqrt{4^2+(-3)^2} \\\\\\ d=\sqrt{16+9}\implies d=\sqrt{25}\implies d=5[/tex]

Answer: second option.

Step-by-step explanation:

You need to use the formula for calculate the distance between two points. This is:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Given the point (1, 4) and the point (5, 1), you can say that:

[tex]x_2=5\\x_1=1\\y_2=1\\y_1=4[/tex]

Now you must substitute these values into the formula.

The distance between the points (1, 4) and (5, 1) is the following:

[tex]d=\sqrt{(5-1)^2+(1-4)^2}\\\\d=5[/tex]

This matches with the second option.

The width of the rectangular field is 139 feet and its length is 320 feet.

To solve this problem, we can use the formula for the perimeter of a rectangle, which is 2*(length + width). We know that the perimeter is 918 feet, and it's given that the length is 181 feet more than the width.

Let's denote the width as w, therefore the length would be w + 181.

By substituting these expressions into the formula for the perimeter, we get:
2*(w + w + 181)=918,
which simplifies to 2*(2w + 181)=918,
further simplifies to 4w + 362 = 918.

To isolate 4w, subtract 362 from both sides:
4w = 918 - 362 = 556.
Finally, divide both sides by 4 to solve for w:
w = 556/4 = 139 feet.

Substitute w = 139 into the length equation to get the length:
length = w + 181 = 139 + 181 = 320 feet.

So the width is 139 feet, and the length is 320 feet.

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Length is 320 feet.

Let's define the width of the rectangle as w feet and the length as l feet.

According to the problem,

the perimeter of the rectangle is 918 feet and the length is 181 feet more than the width.

We can express these conditions using the following equations:

Perimeter equation: 2l + 2w = 918

Length equation: l = w + 181

First, we can solve the perimeter equation for l+w:

2l + 2w = 918
Divide both sides by 2:

l + w = 459

Next, we substitute the length equation l = w + 181 into the perimeter equation:

w + 181 + w = 459 Simplify:

2w + 181 = 459
Subtract 181 from both sides:

2w = 278
Divide by 2:

w = 139 feet

Now substitute the width back into the length equation:

l = w + 181
l = 139 + 181

l = 320 feet

So, the width is 139 feet, and the length is 320 feet.

Let the smaller odd integer be: a

then the larger odd integer which is consecutive to a will be: a+2

It is given that:

Seven times the first (smaller) of two consecutive odd integers is equal to five times the second (larger) integer.

This means that:

7 times of a is equal to 5 times of (a+2)

i.e.

[tex]7a=5(a+2)\\\\i.e.\\\\7a=5\times a+5\times 2[/tex]

( Since, by using the distributive property of multiplication)

i.e.

[tex]7a=5a+10\\\\i.e.\\\\7a-5a=10\\\\i.e.\\\\2a=10\\\\i.e.\\\\a=5[/tex]

Hence, the smaller number is:  5

and the larger number is: 7

( Since a+2=5+2=7 )

Circumference: 18.84 m

Area: 28.26[tex]m^{2}[/tex]

Explanation

For circumference, use the formula [tex]\pi[/tex]*d, where d=the diameter(the radius times 2).

For area, use the formula [tex]\pi[/tex][tex]r^{2}[/tex], where r=the radius. Make sure to square the radius first before multiplying by pi!

A good way to remember these formulas is Cherry(circumference) Pi([tex]\pi[/tex]) is Delicious(diameter) and Apple(area) [tex]\pi[/tex] [tex]r^{2}[/tex]("are too")

The circumference of the circle with radius 3 meters is 18.84 meters and the area is 28.26 square meters.

Given that the radius of the circle is 3 meters, we can find the circumference and the area using the respective formulae.

Circumference of a circle with radius r (C) is given by the formula C = 2πr. Substituting π = 3.14 and r = 3 meters, the circumference C = 2*3.14*3 = 18.84 meters.

The Area of a circle with radius r (A) is given by the formula A = πr². Substituting π = 3.14 and r = 3 meters, the area A = 3.14*3² = 28.26 square meters.

So, the circumference of the circle with radius 3 meters would be 18.84 meters and the area would be 28.26 square meters.

Learn more about circumference here:

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Answer:  There is probability that he will be asked a question in class during Week 7 is 64%.

Step-by-step explanation:

Since we have given that  

Probability that question in class being asked during Week 1 = 1%  

Probability that question in class being asked during Week 2 = 2%  

Probability that question in class being asked during Week 3 = 4%

and so on.

So, we need to find the probability that question being asked in class during week 7.

Since it forms geometric series:

1%, 2%, 4%, ........

So, we need to find the 7 th term:  

[tex]a_n=ar^{n-1}\\\\a_7=ar^{7-1}\\\\a_7=1(2)^6\\\\a_7=64\%[/tex]

Hence, there is probability that he will be asked a question in class during Week 7 is 64%.

Parameterize the first line segment [tex]C_1[/tex] by

[tex]\vec r(t)=(1,0,1)(1-t)+(2,4,1)t=(1+t,4t,1)[/tex]

and the second line segment [tex]C_2[/tex] by

[tex]\vec s(t)=(2,4,1)(1-t)+(2,6,4)t=(2,4+2t,1+3t)[/tex]

both with [tex]0\le t\le1[/tex]. Then

[tex]\displaystyle\int_{C_1}(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz[/tex]

[tex]\displaystyle=\int_0^1\bigg((1+5t)\cdot1+2(1+t)\cdot4+(1-t)4t\cdot0\bigg)\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1(9+13t)\,\mathrm dt=\frac{31}2[/tex]

and

[tex]\displaystyle\int_{C_2}(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz[/tex]

[tex]\displaystyle=\int_0^1\bigg((2+(4+2t)(1+3t))\cdot0+2(2)\cdot2+2(4+2t)(1+3t)\cdot3\bigg)\,\mathrm dt[/tex]

[tex]\displaystyle=\int_0^1(32+84t+36t^2)\,\mathrm dt=86[/tex]

Then

[tex]\displaystyle\int_C(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz=\frac{31}2+86=\boxed{\frac{203}2}[/tex]

To evaluate the line integral, parameterize each segment of the curve separately, substitute the parameterization into the line integral expression, evaluate the integral for each segment, and sum up the results.

To evaluate the line integral, we need to first parameterize the curve C. The given curve consists of two line segments, so we will parameterize each segment separately.

For the first segment from (1, 0, 1) to (2, 4, 1), we can parameterize it as r(t) = (1 + t, 4t, 1), where t ranges from 0 to 1.  

For the second segment from (2, 4, 1) to (2, 6, 4), we can parameterize it as r(t) = (2, 4 + 2t, 1 + 3t), where t ranges from 0 to 1.

Next, we substitute the parameterization into the line integral expression and evaluate the integral for each segment separately. Finally, we sum up the individual results to get the value of the line integral.

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Answer: 0.0325

Step-by-step explanation:

Binomial probability distribution formula for x successes in n trials:-

[tex]P(X=x)=^nC_x\ p^x\ (1-p)^{n-x}[/tex], where n is the number of trials , p is the probability of success.

Given : The probability that taxpayers who filed their tax return are electronically​ self-prepared their taxes : [tex]p= 0.319[/tex]

If three tax returns submitted electronically are randomly​ selected, then the probability that all three were​ self-prepared is given by :-

[tex]P(X=3)=^3C_3\ (0.319)^3\ (1-0.319)^{3-3}\\\\=(1) (0.319)^3(1)\\=0.032461759\approx0.0325[/tex]

Hence, the probability that all three were​ self-prepared =0.0325

Final answer:

The probability that all three tax returns submitted electronically are self-prepared, when each has a 31.9% chance of being self-prepared, is the product of their individual probabilities, resulting in approximately 3.24%.

Explanation:

To compute this, we will use the property that the probability of independent events all occurring is the product of their individual probabilities.

The probability of one taxpayer self-preparing is 0.319. Since the question implies that the taxpayers are chosen independently, the probability that all three returns are self-prepared is:

P(First self-prepared) = 0.319,

P(Second self-prepared) = 0.319,

P(Third self-prepared) = 0.319.

To find the combined probability, we multiply these probabilities together:

P(All three self-prepared) = 0.319 × 0.319 × 0.319,

This calculation results in approximately 0.0324, or 3.24%.

Answer:

  about 12.75%

Step-by-step explanation:

Let r represent the annual rate of return. Compounded annually for 18 years, the account multiplier is (1+r)^18. Then we have ...

  26,000 = 3,000(1+r)^18

  (26,000/3,000)^(1/18) = 1+r . . . . . . divide by 3000, take the 18th root

  (26/3)^(1/18) -1 = r ≈ 12.7465%

Ezio's account earned about 12.75% annually.

Ezio's $3,000 investment grew to $26,000 over 18 years at this rate.

To calculate the annual rate of return Ezio earned on his investment for his son, Flavia's, college tuition, we'll need to use the compound interest formula:

A = P(1 + r)^n

Where:

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

P is the principal amount (the initial amount of money).

r is the annual interest rate (decimal).

n is the number of years the money is invested.

We know that Ezio deposited $3,000 (P = 3000), the amount in the account after 18 years is $26,000 (A = 26000), and that the time period n is 18 years. We need to find the annual interest rate r.

Let's rearrange the formula to solve for r:

(1 + r)^n = A / P

(1 + r)^18 = 26000 / 3000

(1 + r)^18 = 8.6667

Now we need to take the 18th root of 8.6667 to find (1 + r):

1 + r = (8.6667)^(1/18)

The 18th root of 8.6667 is approximately 1.1225. This means:

1 + r = 1.1225

Subtract 1 from both sides to find r:

r = 1.1225 - 1

r = 0.1225 or 12.25%

The annual rate of return Ezio earned on his investment is approximately 12.25%.

Answer:

  9.1%

Step-by-step explanation:

The total number of observations is 33. The number in the second class is 3, so the relative frequency is ...

  3/33 × 100% = (100/11)% ≈ 9.1%

Add everything in the frequency column.

= 33

Take class two (3).

Solution:

3/33 * 100%

1/11 * 100%

100/11%

9.1%

Best of Luck!

Answer:

[tex]f (x) = (x-4)(x + 5)^2(x-9)(x+7)[/tex]

Step-by-step explanation:

The zeros of the polynomial are all the values of x for which the function [tex]f (x) = 0[/tex]

In this case we know that the zeros are:

[tex]x = 4,\ x-4 =0[/tex]

[tex]x = 9,\ x-9=0[/tex]

[tex]x = -5[/tex], [tex]x + 5 = 0[/tex] (multiplicity 2)

[tex]x = -7,\ x+7=0[/tex]

Now we can write the polynomial as a product of its factors

[tex]f (x) = (x-4)(x + 5)^2(x-9)(x+7)[/tex]

Note that the polynomial is of degree 5 because the greatest exponent of the variable x that results from multiplying the factors of f (x) is 5

The polynomial of degree 5 with the zeros 4, 9, -5 (multiplicity 2), and -7 is given by f(x) = (x - 4)(x - 9)(x + 5)² (x + 7).

To find a polynomial f(x) of degree 5 that has the zeros 4, 9, -5 (multiplicity 2), -7, you would begin by setting up a polynomial that has these zeros. From the given roots, we can form the polynomial in factored form as follows:

f(x) = (x - 4)(x - 9)(x + 5)²(x + 7).

Multiplicity refers to how many times a root is repeated. In this case, -5 is repeated twice, hence the exponent of 2.

So, the polynomial in factored form with the zeros provided is:

f(x) = (x - 4)(x - 9)(x + 5)² (x + 7).

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Answer: Our required probability is 0.004672.

Step-by-step explanation:

Since we have given that

Number of adults = 7

Probability of getting adult will watch prime time TV live = 0.8

We need to find the probability that fewer than 3 of the selected adults watch prime time live.

We will use "Binomial Distribution":

here, n = 7

p = 0.8

So, P(X<3)=P(X=0)+P(X=1)+P(X=2)

So, it becomes,

[tex]P(X=0)=(1-0.8)^7=0.2^7=0.0000128[/tex]

[tex]P(X=1)=^7C_1(0.8)(0.2)^6=0.0003584\\\\P(X=2)=^7C_2(0.8)^2(0.2)^5=0.0043[/tex]

So, probability that fewer than 3 of the selected adult watch prime time live is given by

[tex]0.0000128+0.0003584+0.0043=0.004672[/tex]

Hence, our required probability is 0.004672.

This problem relates to the binomial distribution and requires us to find the sum of binomial probabilities for 0, 1, and 2 successes (adults watching live TV) out of seven trials (the seven randomly selected adults).

This question is utilizing the concept of binomial distribution. The probability of a randomly selected adult watching prime-time TV live is 0.8. We want to find the probability that fewer than 3 out of 7 randomly selected adults watch prime-time live.

We find this by adding up the probabilities for 0, 1, and 2 adults watching live TV using the binomial distribution formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)), where C(n, k) denotes the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.

We get:

Adding these up will give the total probability that fewer than three adults out of seven watch prime-time live.

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Answer:

(x+8)^2 + (y-4)^2 = 3^2

or

(x+8)^2 + (y-4)^2 = 9

Step-by-step explanation:

The diameter is 6, so the radius would be d/2 =3

The equation of a circle is

(x-h)^2 + (y-k)^2 = r^2

where (h,k) is the center and r is the radius

(x- -8)^2 + (y-4)^2 = 3^2

(x+8)^2 + (y-4)^2 = 3^2

or

(x+8)^2 + (y-4)^2 = 9

Answer:

The value of [tex]\frac{dx}{dt}[/tex] is [tex]\frac{160}{x^3}[/tex].

Step-by-step explanation:

The given equation is

[tex]x^4=8y^5-240[/tex]

We need to find the value of [tex]\frac{dx}{dt}[/tex].

Differentiate with respect to t.

[tex]4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0[/tex]              [tex][\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0][/tex]

[tex]4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}[/tex]

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

[tex]4x^3\frac{dx}{dt}=40(2)^4(1)[/tex]

[tex]4x^3\frac{dx}{dt}=40(16)(1)[/tex]

[tex]4x^3\frac{dx}{dt}=640[/tex]

Divide both sides by 4x³.

[tex]\frac{dx}{dt}=\frac{640}{4x^3}[/tex]

[tex]\frac{dx}{dt}=\frac{160}{x^3}[/tex]

Therefore the value of [tex]\frac{dx}{dt}[/tex] is [tex]\frac{160}{x^3}[/tex].

Answer:

  B.  AAS Congruence Theorem

Step-by-step explanation:

Previous steps showed congruence of a side of the designated triangle, and two angles that do not bracket that side. In short form, you have shown congruence of ...

  Angle - Angle - Side

so the AAS congruence theorem applies.

Step-by-step answer:

Given:

class time uniformly distributed between 48 and 53 minutes (range 5 minutes)

Find probability that the class period runs between 51.5 and 51.75 minutes.

Solution:

Uniformly distributed means that the probability that the class time is the same for any minute (out of 5), i.e. 20%, or probability density is 0.2 / minutes, no matter which minute.

The interval between 51.5 and 51.75 is 0.25 minutes, so the probability of class period within that range is 0.25 minutes * 0.2 / minute = 0.05.

Final answer:

The probability that a class period runs between 51.5 and 51.75 minutes, when class lengths are uniformly distributed between 48.0 and 53.0 minutes, is 0.05 or 5%.

Explanation:

To find the probability that a given class period runs between 51.5 and 51.75 minutes when class lengths are uniformly distributed between 48.0 and 53.0 minutes, we utilize the properties of the uniform distribution. The probability is the area under the uniform probability density function (PDF) between the two specified lengths. For a uniform distribution, this area corresponds to the length of the interval divided by the length of the entire interval in which the variable is uniformly distributed.

Here, we're looking for the probability between 51.5 minutes and 51.75 minutes. The total length of the available interval is 53.0 - 48.0 = 5 minutes. The length of the desired interval is 51.75 - 51.5 = 0.25 minutes. So, we divide the length of the desired interval by the total interval to get the probability.

Probability = (51.75 - 51.5) / (53.0 - 48.0) = 0.25 / 5 = 0.05

Therefore, the probability that a class period runs between 51.5 and 51.75 minutes is 0.05 or 5%.

1. (1010 0110)₂ = (166)₁₀

2. (145)₁₀ = (1001 0001)₂

3. (101 000 111 100)₂ = (5074)₈

4. (1111 1100 0011 0110)₂ = (FC36)₁₆

1. Convert the 8-binary binary expansion (1010 0110)₂ to a decimal expansion.

In order to solve this problem we have to use the expansion:

n = aₓbˣ + aₓ₋₁bˣ⁻¹ + ... + a₁b¹ + a₀

where b = 2, x = 8 - 1 = 7 due is a 8-binary

(1010 0110)₂ = 1 x 2⁷ + 0 x 2⁶ + 1 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2⁰

(1010 0110)₂ = 128 + 0 + 32 + 0 + 0 + 4 + 2 + 0

(1010 0110)₂ = (166)₁₀

2. Convert the following decimal expansion (145)₁₀ to an 8-bit binary expansion.

To solve this problem we have to use the divide by 2 process.

Since we are dividing by 2, when the dividend is an even number, the remainder will be 0, and when the dividend is an odd number the binary residual will be 1.

145 ---------------> 1  Less significant bit

145/2 = 72 -----> 0

72/ 2 = 36 -----> 0

36/2 = 18 ------> 0

18/2 = 9 -------> 1

9/2 = 4 --------> 0

4/2 = 2 ---------> 0

2/2 = 1 ----------> 1  Most significant bit

Then we order from the most significant bit to the less significant bit (from the bottom to the top) to obtain the 8-binary number:

(145)₁₀ = (1001 0001)₂

3. Convert the following hexadecimal expansion (A3C)₁₆ to an octal expansion.

To convert a hexadecimal expansion to an octal expansion we have to convert  from hexadecimal to binary and then to octal using the table hexadecimal to binary and binary to octal.

Converting from hexadecimal to binary:

(A3C)₁₆

A = 1010, 3 = 0011 and C = 1100

(A3C)₁₆ = (1010 0011 1100)₂

Converting from binary to octal:

To convert binary to octal we have to order the binary expansion into group of 3-bits and use the table to convert binary to octal.

(1010 0011 1100)₂ = (101 000 111 100)₂

101 = 5, 000 = 0, 111 = 7 and 100 = 4

(101 000 111 100)₂ = (5074)₈

4. Convert the following binary expansion (1111 1100 0011 0110)₂ to a hexadecimal expansion.

To solve this exercise we have to use the binary to hexadecimal table.

(1111 1100 0011 0110)₂

1111 = F, 1100 = C, 0011 = 3 and 0110 = 6

(1111 1100 0011 0110)₂ = (FC36)₁₆

Step-by-step explanation:

Consider the provided information.

Jason received an invoice for $5,000, and the invoice was dated on August 26.

Part 1)

We need to find the bill Jason needs to pay by September 26.

n/45 means invoice amount should be paid within 45 days after the august 26.

From August 26 to September 26 Jason complete a month.

Thus, Jason just need to pay only $5,000. Which is the invoice amount.

Part 2)

The last day of discount can be calculated by 2/10.

Here, the 2/10 represents that discount of 2% within 10 days.

Now, calculate the 10 days form August 26.

The 10 days from August 26 is September 5.

Hence, September 5 is the last day that Jason can receive the cash discount.

Part 3)

Now, we need to calculate the last day that Jason has to pay for the bill.

n/45 means invoice amount should be paid within 45 days after the august 26.

Hence, the sum of the invoice should be paid within 45 days of August 26, whatever the date.

Answer:

The total cost to install the 1,200 ft of the pipe is $21000.

Step-by-step explanation:

Consider the provided information.

You buy 900 ft pipe with a cost of $15 per foot and the remaining 300 ft pipe with a cost of $25 per foot.

Therefore, the total cost is:

[tex]900 \times 15+300 \times 25[/tex]

[tex]13500+7500[/tex]

[tex]13500+7500[/tex]

[tex]21000[/tex]

Hence, the total cost to install the 1,200 ft of the pipe is $21000.

Answer:

c > 9

Step-by-step explanation:

Isolate the variable, c. Treat the greater than sign as an equal, what you do to one side, you do to the other. Add 7 to both sides:

c - 7 > 2

c - 7 (+7) > 2 (+7)

c > 2 + 7

c > 9

c > 9 is your answer.

~

Answer:

[tex]\Huge \boxed{C>9}\checkmark[/tex]

Step-by-step explanation:

Add by 7 from both sides.

[tex]\displaystyle c-7+7>2+7[/tex]

Simplify, to find the answer.

[tex]\displaystyle 2+7=9[/tex]

[tex]\huge \boxed{c>9}[/tex], which is our answer.

Answer: 0.128

Step-by-step explanation:

The geometric probability of getting success on nth trial is given by :-

[tex]P(n)=p(1-p)^{n-1}[/tex]

Given : The probability that he is successful on a given trial[tex]p= 0.2[/tex].

Then , the probability that the third hole drilled is the first to yield a productive well is given by :-

[tex]P(3)=0.2(1-0.2)^{3-1}=0.128[/tex]

Hence, the probability that the third hole drilled is the first to yield a productive well = 0.128

Answer:

it would be A 1251

Step-by-step explanation:

The question involves hypothesis testing of proportions. Based on a sample where 80% favour Candidate A, we may infer the true population proportion is significantly greater than 75%, however, actual mathematical calculations are needed for confirmation.

The question examines whether the proportion of the population favouring Candidate A is significantly more than 75% based on a sample of 100 people where 80 favoured Candidate A. This is a problem of hypothesis testing for proportions. The null hypothesis (H₀) is that the true population proportion is 75% (p = 0.75), versus the alternative hypothesis (H₁) stating the true population proportion is more than 75%. Using a significance level of 0.05, we examine the data.

With a sample proportion of 80 out of 100 (p' = 0.80) and given the large sample size, we apply the Normal approximation to the Binomial distribution, followed by a one-sample z-test. If the resulting p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the true population proportion of individuals favouring Candidate A is significantly greater than 75%.

However, without performing the actual calculations, we cannot definitively determine the conclusion. From a practical perspective, an 80% sample proportion showing favour in a sample as large as 100 might indicate a significantly higher proportion than 75%, but an exact mathematical test should be done to confirm this.

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Answer:

0.0488%

Step-by-step explanation:

Here we have the probability of different independent events, which means that none of the previous ones affect the next. For this kind of events, the probability 'P' that a series of events occur is the multiplication of the probability of each singular event.

p1: Probability that an H be obtained: 50% (is always 50% because it is independent of the previous results)

P: The probability that H be obtained all of the 10 times. This is the complementary probability to E:(a T comes up at least once).

By the first definition given

[tex]P=0.5^{11}=0.00048828=0.0488%[/tex]

The complementary probability 'P' is the probability that 'E' does not happen, so the probability that E happen is: [tex]P_{E} =1-P[/tex]

The last makes sense if we think about the fact that for the experiment there are just two possibilities, 'E' happen, or 'E' does not happen. Then,

[tex]P_{E} =1-0.000488=99.95[/tex]%

Answer:

  2280 kcal

Step-by-step explanation:

The given function R(t) is periodic with a period of 24 hours, so the integral is the product of the average value (95 kcal/h) and the 24-hour interval:

  (95 kcal/h)(24 h) = 2280 kcal

To calculate the total basal metabolism over 24 hours for a young man modeled by R(t) = 95 − 0.18 cos(πt/12), you need to apply integral calculus. First, integrate the function R(t) and then apply a definite integral over the period from 0 to 24 hours. The result will be the total basal metabolism over this time period.

To solve this problem, you need to integrate the function R(t) = 95 − 0.18 cos(πt/12) over the time interval of 0 to 24 hours. In calculus, such an operation is represented as ∫R(t) dt from 0 to 24. This operation will give you the total basal metabolism during these 24 hours.

The first thing you need to do in this case is to integrate the function. As this function is a composition of functions, you would need to use the basics of integral calculus, specifically the trigonometric substitution method or table of integrals for cosine function.

Afterward, apply a definite integral over the period from 0 to 24 hours. This will give you the total basal metabolism for 24 hours, expressed in kcal.

The steps presented above represent a simplification of the process. The actual calculations might be quite complex, so please don't worry if they seem overwhelming - it's a normal part of learning integral calculus.

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Answer with explanation:

It is given that , a and b are positive integers.

gcd(a,b)=1

We have to prove for any positive integer x and y ,

a x + by =c, for any integer greater than ab-a-b.

Proof:

GCD of two numbers is 1, when two numbers are coprime.

Consider two numbers , 9 and 7

GCD (9,7)=1

So, we have to calculate positive integers x and y such that

⇒ 9 x +7 y > 9×7-9-7

⇒9x +7 y> 47

To prove this we will draw the graph of Inequality.

So the ordered pair of Integers are

x>5 and y>6.

So, for any integers , a and b ,

→ax+ by > a b -a -b, if

[tex]\Rightarrow \frac{x}{\frac{ab-a-b}{a}}+ \frac{y}{\frac{ab-a-b}{b}}>1,\frac{ab-a-b}{a},\text{and},\frac{ab-a-b}{b}[/tex]

⇒Range of x for which this inequality hold

      [tex]=[\frac{ab-a-b}{a},\infty)[/tex]

if,

[tex]\frac{ab-a-b}{a}[/tex]

is an Integer ,otherwise range of x

         [tex]=(\frac{ab-a-b}{a},\infty)[/tex]

⇒Range of y for which this inequality hold

      [tex]=[\frac{ab-a-b}{b},\infty)[/tex]

if,

[tex]\frac{ab-a-b}{b}[/tex]

is an Integer ,otherwise range of y

         [tex]=(\frac{ab-a-b}{b},\infty)[/tex]

Answer: [tex]H_0:\mu\geq431\\\\H_a:\mu<431[/tex]

Step-by-step explanation:

Given : A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 431 gram setting. t is believed that the machine is underfilling the bags.

Claim : [tex]\mu<431[/tex]

We know that the null hypothesis always have equals sign , then the required set of hypothesis for the situation :-

[tex]H_0:\mu\geq431\\\\H_a:\mu<431[/tex]

Answer:

a) 0.8508

b) 0.1389

c) 0.0099

Step-by-step explanation:

Total number of calculators bought = 100

Number of calculators with defect = 2

Probability of selecting a calculator with defect = p = 2 out of 100 = [tex]\frac{2}{100}=0.02[/tex]

Probability of selecting a calculator without defect = q = 1 - p = 1 - 0.02 = 0.98

Part a)

The bookstore buys 8 calculators. This means the number of trials is fixed ( n =8). The calculator is either with defect or without defect. This means there are 2 outcomes only. The outcome of one selection is independent of other selections. This means all the events(selections) are independent of each other.

Hence, all the conditions of a Binomial Experiment are being satisfied. So we will use Binomial Probability to solve this problem.

We need to find the probability that all calculators are free of defects. i.e. number of success is 0 i.e. P( x = 0 )

The formula of Binomial Probability is:

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

Using the values, we get:

[tex]P(0)=^{8}C_{0}(0.02)^{0}(0.98)^{8-0}\\\\ P(0)=0.8508[/tex]

Thus, the probability that all calculators are free of defects is 0.8508

Part b) Exactly 1 calculator will have defect

This means the number of success is 1. So, we need to calculate ( x = 1)

Using the formula of binomial probability again and using x = 1, we get:

[tex]P(1)=^{8}C_{1}(0.02)^{1}(0.98)^{8-1}\\\\ P(1)=0.1389[/tex]

Thus, the probability that exactly two calculators will be having a defect is 0.1389

Part c) Exactly 2 calculator will have defect

This means the number of success is 2. So, we need to calculate ( x = 2 )

Using the formula of binomial probability again and using x = 2, we get:

[tex]P(2)=^{8}C_{2}(0.02)^{2}(0.98)^{8-2}\\\\ P(2)=0.0099[/tex]

Thus, the probability that exactly two calculators will be having a defect is 0.0099

Answer:

[tex]y=c_1x+c_2x^5[/tex]

Step-by-step explanation:

The given second order homogeneous Cauchy-Euler ordinary differential equation is

[tex]x^2y''-5xy'+5y=0[/tex]

The corresponding auxiliary equation is given by:

[tex]a {m}^{2} + (b - a)m + c = 0[/tex]

where a=1, b=-5, c=5

We substitute the coefficients into the auxiliary equation to obtain:

[tex] {m}^{2} + ( - 5 - 1)m + 5= 0[/tex]

[tex] {m}^{2} - 6m + 5= 0[/tex]

[tex](m - 1)(m - 5) = 0[/tex]

[tex] \implies \: m = 1 \: or \: m = 5[/tex]

The auxiliary equation has two distinct real roots. The general solution to the corresponding differential equation is of the form:

[tex]y=c_1x ^{m1} +c_2 {x}^{m2} [/tex]

We substitute the values to get:

[tex]y=c_1x+c_2x^5[/tex]

The provided differential equation [tex]x^2y'' - 5xy' + 5y = 0[/tex] can be solved using the method for Cauchy-Euler equations, resulting in the general solution [tex]y = c_1 x + c_2 x^5.[/tex] Therefore, the correct answer is (d) [tex]y = c_1 x^{5 + \frac{\sqrt{5}}{2}} + c_2 x^{5 - \frac{\sqrt{5}}{2}}[/tex]

To solve the given differential equation [tex]x^2y'' - 5xy' + 5y = 0[/tex], we use the method for Cauchy-Euler equations:

1. Assume a solution of the form[tex]y = x^m.[/tex]

2. Substitute [tex]y = x^m[/tex] into the differential equation:

3. Plug these into the equation:
[tex]x^2 [m(m-1)x^(m-2)] - 5x[mx^(m-1)] + 5[x^m] = 0[/tex]

4. Simplify to get the characteristic equation:
[tex]m(m-1) - 5m + 5 = 0[/tex]
[tex]m^2 - 6m + 5 = 0[/tex]

5. Solve the quadratic equation:
[tex]m = \frac{{6 \pm \sqrt{6^2 - 4 \times 5}}}{{2}}[/tex]
[tex]m = \frac{{6 \pm \sqrt{16}}}{{2}}[/tex]
[tex]m = \frac{{6 \pm 4}}{2}[/tex]
[tex]m = 5 \quad \text{or} \quad m = 1[/tex]

6. The general solution is then a linear combination of these solutions:
[tex]y = c_1 x^5 + c_2 x^1[/tex]
[tex]y = c_1 x + c_2 x^5[/tex]

[tex]y = c_1 x^{5 + \frac{\sqrt{5}}{2}} + c_2 x^{5 - \frac{\sqrt{5}}{2}}[/tex]

Answer:

Viatical settlement

Step-by-step explanation:

Here Barney sold his life insurance to 3rd party for less net death benefit value.

He entered into the arrangement called - viatical settlement

A viatical settlement is the sale of a life insurance policy to a third party. The owner of the life insurance policy sells the policy for cash benefit, thus making the buyer the new owner of the policy who will get death benefits.