anyone have free time to answer

Answer:

The height (corresponding to the [tex] \\ 92^{nd}[/tex] percentile) is (to the nearest inch) 73 inches (and, approximately, only 8% of adult males are taller than this height.)

Step-by-step explanation:

Roughly speaking, the [tex] \\ 92^{nd}[/tex] percentile is the x value (in the distribution) for which 92% of the observations in the [normal] distribution are below this x value, and 8% of the observations are above this x value.

To answer this question, we already know that:

Strategy for solving the question

For solving this, we have to use here the following key concepts: z-scores, the cumulative standard normal distribution, and the cumulative standard normal table.

Z-scores

To find the [tex] \\ 92^{nd}[/tex] percentile, we can use z-scores or standardized values. A z-score is a value that tells us the distance in standard deviations units from the mean. When the z-score is positive, it means that the value is above the mean. A negative indicates that the z-score is below the mean. The formula to obtain a z-score is as follows:

[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]

Where

Cumulative standard normal distribution and corresponding table

We still need to know the corresponding z-score, z, for the cumulative probability of 92%. For this, we have to consult the standard normal table, available on the Internet or in any Statistics books.

In this case, we look in the different columns of the standard normal table a probability value (exact or approximate) to 0.92 and then find the value for z that corresponds to this probability. The value for z is between 1.40 (0.91924) and 1.41 (0.92073).

Using z = 1.40 in [1], we have:

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

[tex] \\ 1.40 = \frac{x - 69}{3}[/tex]

Then, solving for x:

Multiplying by 3 at each side of the equation:

[tex] \\ 1.40 * 3 = x - 69[/tex]

Adding 69 at both sides of the equation:

[tex] \\ (1.40 * 3) + 69 = x[/tex]

[tex] \\ x = (1.40 * 3) + 69[/tex]

[tex] \\ x = 4.20 + 69[/tex]

[tex] \\ x = 73.20[/tex]

That is, the [tex] \\ 92^{nd}[/tex] percentile is 73.20 inches, and to the nearest inch, this percentile is 73 inches.

This result indicates that, approximately, 92% of the heights are below 73 inches, and only 8% of heights are taller than this height.  

The shaded area in the graph below shows an area of 0.08076 (8.076%) for 73.20 inches.  

Answer:

Jomo Kenyatta

Step-by-step explanation:

Jomo Kenyatta was a Kenyan politician, who was one of the first African anti-colonial figures. He became the prime minister of Kenya from 1963 to 1964, and after Kenyan independence in 1964, he became president of Kenya. Jomo Kenyatta was born into a family of Kikuyu farmers in Kiambu, present day Kenya which was then, British East Africa. He had his basic schooling in a missionary school before proceeding to study at Moscow's Communist University of the Toilers of the East, University College London, and the London School of Economics.

Answer:

Jomo Kenyatta

Step-by-step explanation:

took the test

Answer:

Increase in age is indirectly proportional to anger intensity in children's tantrums

Step-by-step explanation:

Let's begin by reproducing the given information into a frequency distribution table (shown below):

Age Group   Frequency

   0 - <2              136

   2 - <4              92

   4 - <11              72

   11 - <20           26

   20 - <30          7

   30 - <40          3

Observing the dataset above, we will see a trend; increase in age is inversely proportional to anger intensity. Kindly note that the histogram is attached as a picture.

Analysis of Histogram

Age group 1 (infants) have the highest indices of anger intensity in tantrums. This is expected because they do not know much yet and have an entitlement mentality.

Age group 2 (infants) also have a pretty high amount of anger intensity in children tantrums. They have transitioned from infancy but are yet immature & can easily flare up when disgruntled.

Age group 3 (children) are not far off from the previous age bracket, also having a significantly fairly high anger intensity. They are advancing in years & as they do, their emotional rage to tantrums is reducing.

Age group 4 (pre-teens/teens) experience a significant and drastic drop in emotional rage due to children's tantrums. They are maturing and are learning to communicate in less fitful ways. They now know better than to express their displeasure like infants.

Age group 5 (youths) also see a drastic drop in such anger intensity. The reason is not far-fetched; they are more mature and can express their displeasure in more mature ways.

Age group 6 (Adults) almost have almost a zero or non-existent such attitude among them. At this point, they have come to understand & know that they can get much more done with words than such outlandish methods.

In conclusion, as a child advances in years (changing from one age group to another → from infant to adult), their anger intensity takes a nosedive. We rightly interpret the histogram when we say that physical maturity is inversely proportional to anger intensity in children's tantrums.

Step-by-step explanation:

Let y1 and y2 be (e^x)/2, and (xe^x)/2 respectively.

The Wronskian of them functions be

W = (y1y2' - y1'y2)

y1 = (e^x)/2 = y1'

y2 = (xe^x)/2

y2' = (1/2)(x + 1)e^x

W = (1/4)(x + 1)e^(2x) - (1/4)xe^(2x)

= (1/4)(x + 1 - x)e^(2x)

W = (1/4)e^(2x)

Since the Wronskian ≠ 0, we conclude that functions are linearly independent, and hence, form a set of fundamental solutions.

Answer:

W = (1/4)(x + 1)e^(2x) - (1/4)xe^(2x)

= (1/4)(x + 1 - x)e^(2x)

W = (1/4)e^(2x)

Step-by-step explanation:

Answer:

The coordinates of the new position are

(x, y, z) = (7, 0, -9)

Step-by-step explanation:

So assuming a 3D plane x, y, z

The starting point is the origin that is

(x, y, z) = (0, 0, 0)

Move a distance of 7 units along the positive x-axis

(x, y, z) = (7, 0, 0)

Then move a distance of 9 units downwards along the z-axis

downward = negative and upward = positive

(x, y, z) = (7, 0, -9)

Therefore, the coordinates of the new position are

(x, y, z) = (7, 0, -9)

Refer to the attached plot where the above coordinates are plotted on a 3D surface using an online 3D plotter.

Each small box corresponds to 1 unit.

p-value < 0.05, reject null hypothesis and we can conclude that data is not consistent with distributor's claim.

Given, data of toiletry distributors.

Total number of observations

= 170 + 105 + 80 + 45

= 400

Expected count = [tex]p_i \times 400[/tex]

Calculation table is attached below.

Test statistic:

Chi square score

= 6.429 + 0.438 + 0.000 + 7.779

= 14.645

Degree of freedom:

df = 4-1

=3

p-value = CHIDIST(14.645,3)

= 0.00215

Therefore , p-value < 0.05

Reject null hypothesis and we can conclude that data is not consistent with distributor's claim.

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Answer:

2.87 = AC

Step-by-step explanation:

Since this is a right triangle we can use trig functions

sin theta = opp / hyp

sin 35 = AC /5

5 sin 35 = AC

2.867882182= AC

To the nearest hundredth

2.87 = AC

Answer:

C. The 4 columns of UQ list the total costs for materials, labor, and overhead used to manufacture products B and C during the 4 quarters of the year.

Step-by-step explanation:

[tex]U=\left\begin{array}{ccc}\\$Cost of Materials\\ $Cost of Labour\\ $Cost of Overhead\end{array}\right\left(\begin{array}{cc}$Product B&$Product C\\0.45&0.42\\ 0.25&0.35\\ 0.15&0.15\end{array}\right)[/tex]

[tex]q_1[/tex] is a vector in the set of real numbers [tex]R^2[/tex] that lists the output​ (measured in​ dollars) of products B and C manufactured during the first quarter of the​ year.

Therefore:

[tex]UQ=\left\begin{array}{ccc}\\$Cost of Materials\\ $Cost of Labour\\ $Cost of Overhead\end{array}\right\left(\begin{array}{cc}$Product B&$Product C\\0.45&0.42\\ 0.25&0.35\\ 0.15&0.15\end{array}\right)\left(\begin{array}{ccc}q_{1B}\\q_{1C}\end{array}\right)\left(\begin{array}{ccc}q_{2B}\\q_{2C}\end{array}\right)\left(\begin{array}{ccc}q_{3B}\\q_{3C}\end{array}\right)\left(\begin{array}{ccc}q_{4B}\\q_{4C}\end{array}\right)[/tex]

[tex]=\left(\begin{array}{c|c|c|c}q_1&q_2&q_3&q_4\\0.45q_{1B}+0.42q_{1C}&0.45q_{2B}+0.42q_{2C}&0.45q_{3B}+0.42q_{3C}&0.45q_{4B}+0.42q_{4C}\\0.25q_{1B}+0.35q_{1C}&0.25q_{2B}+0.35q_{2C}&0.25q_{3B}+0.35q_{3C}&0.25q_{4B}+0.35q_{4C}\\0.15q_{1B}+0.15q_{1C}&0.15q_{2B}+0.15q_{2C}&0.15q_{3B}+0.15q_{3C}&0.15q_{4B}+0.15q_{4C}\end{array}\right)[/tex]Therefore, UQ has 4 columns and 3 rows.

The 4 columns of UQ list the total costs for materials(Row 1), labor(Row 2), and overhead(Row 3) used to manufacture products B and C during the 4 quarters of the year.

Answer:

1 1/6

Step-by-step explanation:

Edwin used a quantity of 2.833 gallons of paint for the mural which is the difference between the quantity of paint at the beginning and the used for the bedroom.

We have been given that Edwin has 3 1/2 gallons of green paint. He uses 2/3 of the paint to paint his bedroom. He uses the rest of the paint for a mural in his garage.

A fraction is defined as a numerical representation of a part of a whole that represents a rational number.

To determine the quantity of paint Edwin used for the mural.

The quantity of paint Edwin used for the mural is the difference between the quantity of paint at the beginning and the used for the bedroom.

The quantity of paint used for the mural = 3 1/2 gallons - 2/3 gallons

The quantity of paint used for the mural = 3.5 - 0.66

The quantity of paint used for the mural = 2.833

Thus, Edwin used a quantity of 2.833 gallons of paint for the mural.

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Answer:

The function is exponential.

The function increases by a factor of 2.5 for each unit increase in x.

The domain of the function is all real numbers

The true statements are:

The function is given as:

[tex]\mathbf{y = 3(2.5)^x}[/tex]

An exponential function is represented as:

[tex]\mathbf{y = ab^x}[/tex]

Where: a represents the initial value, and b represents the rate

This means that:

[tex]\mathbf{y = 3(2.5)^x}[/tex] is an exponential function

By comparison:

[tex]\mathbf{b = 2.5}[/tex]

When b > 0, then the function represents an exponential growth

2.5 is greater than 0.

So, the function represents an exponential growth

Lastly, there is no restriction to the values of x.

So, the domain of the function is the set of all real numbers

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Work Shown:

The green triangle in the back has height 2.6 and an unknown base x. Half of this is x/2, which I'll call y. So y = x/2.

The green triangle in the back is split along the vertical dotted line to get two right triangles. The base of each right triangle is y = x/2.

Use the Pythagorean theorem to find y. Use that to find x

a^2+b^2 = c^2

y^2+(2.6)^2 = (3.2)^2

y^2 + 6.76 = 10.24

y^2 = 10.24 - 6.76

y^2 = 3.48

y = sqrt(3.48) .... apply square root

y = 1.8654758 approximately

x/2 = 1.8654758

x = 2*1.8654758

x = 3.7309516 also approximate

The base of the triangle is roughly 3.7309516 meters

We can now find the area of one green triangle

area of triangle = base*height/2 = 3.7309516*2.6/2 = 4.85023708

two triangles have approximate area 2*(4.85023708) = 9.70047416

----------------------------------

So far we've only considered the triangular faces. There are 3 more faces which are the rectangular sides. These are known as the lateral sides.

One way to get the lateral surface area is to multiply the perimeter of the triangle by the depth of the prism

perimeter of triangle = (side1)+(side2)+(side3)

perimeter = 3.7309516 + 3.2 + 3.2

perimeter = 10.1309516

lateral surface area = (depth)*(perimeter)

lateral surface area = (8.26)*(10.1309516)

lateral surface area = 83.681660216

----------------------------------

The last step is to add this lateral surface area onto the area of the two triangles to get the full surface area

surface area = (triangular area) + (lateral surface area)

surface area = (9.70047416) + (83.681660216)

surface area = 93.382134376

surface area = 93.382 square cm

Complete question is;

A utility company offers a lifeline rate to any household whose electricity usage falls below 240 kWh during a particular month. Let A denote the event that a randomly selected household in a certain community does not exceed the lifeline usage during January, and let B be the analogous event for the month of July (A and B refer to the same household). Suppose P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8. Compute the following.

a. P(A ∩ B).

b. The probability that the lifeline usage amount is exceeded in exactly one of the two months.

Answer:

A) 0.4

B) 0.4

Step-by-step explanation:

We are given;

P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8

A) To solve this question, we will use the the general probability addition rule for the union of two events which is;

P(A∪B) = P(A) + P(B) − P(A∩B)

Making P(A∩B) the subject of the equation, we have;

P(A∩B) = P(A) + P(B) − P(A∪B)

Thus, plugging in the relevant values, we have;

P(A∩B) = 0.7 + 0.5 - 0.8

P(A∩B) = 0.4

B)The probability that the lifeline usage amount is exceeded in exactly one of the two months can be described in terms of A and B as:

P(A but not B) + P(B but not A) = P(A∩B') + P(B∩A')

where;

A' is compliment of set A

B' is compliment of set B

Now,

P(A∩B') = 0.7 − 0.4 = 0.3

P(B∩A') = 0.5 − 0.4 = 0.1

Thus;

P(A but not B) + P(B but not A) = 0.1 + 0.3 = 0.4

Answer:

No. If a collegue hire 7 math Ph.D.s, at least 2 will be from the same university.

The only way to have a job for all of them is that all colleges hire 6 math Ph.D.s, one from each university.

Step-by-step explanation:

If the condition is that no college will hire more than one Ph.D. from any given university, no college can hire more than 6 math Ph.D.s, as there are only 6 universities to choose from. If they hire seven, they have to repeat some university, violating the condition.

Then, the only way to have a job for all of them is that all colleges hire 6 math Ph.D.s, one from each university.

Answer:

Monto total de inicio requerido, de acuerdo con los detalles proporcionados en la pregunta = $ 1,264,000

Total start-up amount required, according to the details provided in the question = $1,264,000

Step-by-step explanation:

- Hay 1000 m² de espacio de almacén para construir a $ 42 por m². Dinero total requerido = 1000 × 42 = $ 42,000.

- Hay 6000 m² de espacio de acceso pavimentado para construir a $ 32 por m². Dinero total requerido = 6000 × 32 = $ 192,000.

- Compra de servidores de última generación para revisar los autobuses antes de comenzar sus recorridos. Costo total = $ 630,000.

- Se necesita comprar equipo especial para revisar los neumáticos. Costo = $ 400,000.

Monto total de inicio requerido, de acuerdo con los detalles proporcionados en la pregunta = 42000 + 192000 + 630000 + 400000 = $ 1,264,000

¡¡¡Espero que esto ayude!!!

English Translation

- There is 1000 m² of warehouse space to construct at $42 per m². Total required money = 1000 × 42 = $42,000

- There is 6000 m² of paved access space to construct at $32 per m². Total money required = 6000 × 32 = $192,000

- Purchase of state-of-the-art servers to check the buses before starting their tours. Total Cost = $630,000

- Purchase of special equipment is needed to check the tires. Cost = $400,000

Total start-up amount required, according to the details provided in the question = 42000 + 192000 + 630000 + 400000 = $1,264,000

Hope this Helps!!!

Answer:8 i ithink

Step-by-step explanation:

Answer:

12, go for 24.

Step-by-step explanation:

There are 6 sides of a cube.

There are 2 pairs of parallel line segments for each side.

6 x 2 = 12

Although that answer is not there, you should go for 24. Since there are 2 variables for each line segment, 12 x 2 = 24. Not sure, hope this helps.

:/

Answer:

[tex]k \approx 0.44[/tex]

Step-by-step explanation:

Given function:

[tex]f(t) = (ce)^{-kt}+32[/tex]

As per question statement:

Initial temperature of bottle is 70 [tex]^\circ F[/tex].

i.e. when time = 0 minutes, f(t) = 70 [tex]^\circ F[/tex]

[tex]70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38[/tex]

After t = 3, f(t) = 42[tex]^\circ F[/tex]

[tex]42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44[/tex]

Hence, the value is:

[tex]k \approx 0.44[/tex]

Answer: 35%

Step-by-step explanation:

If no is 10, 10 x 0.65 = 6.5. OR

10 - 35% of 10 = 6.5

  Multiplying by 0.65 is the same as decreasing by 35%

Let the number is 'a' and percentage decrease is 'b',

Expression for the given statement will be,

a × 0.65 = a - (b% of a)

[tex]0.65a=a(1-\frac{b}{100})[/tex]

[tex]0.65=1-\frac{b}{100}[/tex]

[tex]\frac{b}{100}=1-0.65[/tex]

[tex]b=100(0.35)[/tex]

[tex]b=35[/tex]

    Therefore, Multiplying by 0.65 is the same as decreasing by 35%.

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Answer:

N(L) = 20

The expected number of left handed people is 20.

Step-by-step explanation:

Given;

Percentage of left handed people P(L) = 10%

Total number of selected people N(T) = 200

The Expected number of left handed people N(L) is;

N(L) = Total number of selected people × Percentage of left handed people/100%

N(L) = N(T) × P(L)/100%

Substituting the given values;

N(L) = 200 × 10%/100%

N(L) = 200 × 0.1

N(L) = 20

The expected number of left handed people is 20.

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

n = 10

Compute the sample mean and sample standard deviation:

[tex]\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7[/tex]

[tex]s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702[/tex]

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a t confidence interval.

The critical value of t for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

[tex]t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262[/tex]

*Use a t-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}[/tex]

     [tex]=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)[/tex]

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Answer:

AT least 14 classrooms to hold the total number of students

Step-by-step explanation:

Since  we don't know the numer of girls in the school, let's call it "x".

What we know is that the number of girls plus the number of boys gives the total number of students. This is:

x + 129 = Total number of students

Now, since 5/8 of the total number of students are girls, and understanding that 5/8 = 0.625 in decimal form, then we write the equation that states:

"5/8 of the school's population are girls" as:

0.625 (x + 129) = x

then we solve for "x":

0.625 x + 0.625 * 129 = x

0.625 * 129 = x - 0.625 x

80.625 = x (1 - 0.625)

80.625 = 0.375 x

x = 80.625/0.375

x = 215

So now we know that the total number of students is: 215 + 129 = 344

and if each classroom can hold 25 students, the number of classroom needed for 344 students is:

344/25 = 13.76

so at least 14 classrooms to hold all those students

Answer:

A shape with two pairs of parallel lines, perpendicular lines, and two pairs of equal sides can be best described as a rectangle.

Answer:

Index numbers are intended to measure the degree of economic changes over time. These numbers are values stated as a percentage of a single base figure. Index numbers are important in economic statistics. ... Index numbers are intended to study the change in the effects of such factors which cannot be measured directly.

Answer:Index numbers are important in economic statistics. In simple terms, an index (or index number) is a number displaying the level of a variable relative to its level (set equal to 100) in a given base period. Index numbers are intended to study the change in the effects of such factors which cannot be measured directly.

Step-by-step explanation:

Answer:

False

Step-by-step explanation:

from *millermoldwarp*

"Events are called dependent when the probability of an event depends on the occurrence of another. When event A depends on event B, the probability that A occurs, given that B has occurred, is different from the probability that A occurs only ."

hopes this helps

Answer:false

Step-by-step explanation:

Answer:

Step-by-step explanation:

Mean is defined as the sum total of data divided by the total number of data.

[tex]\overline x = \frac{\sum Xi}{N}[/tex]

Xi are individual data

N is the total number of data = 5

Given the mean of the pulse rate = 65.4

a) Let the 5pulse rates be our data as shown: 57​, 55​, 62​, 83 and y where y is the missing value. According to the formula:

[tex]65.4 = \frac{57+55+62+83+y}{5}[/tex]

[tex]65.4 = \frac{257+y}{5}\\ 257+y = 65.4*5\\257+y = 327\\y = 327-257\\y = 70[/tex]

The missing value is 70

b) Since the total list of numbers is n values with a specific known mean, if some of this values can be freely selected, the number of n values that can be freely assigned before the remaining values are​ determined is any values less than n i.e n-1 values.

From the formula for calculating mean:

[tex]\overline x = \frac{\sum Xi}{N}\\{\sum Xi} = N\overline x[/tex]

This shows that the sum of all the values is equal to the product of the total values and the mean value. Since we can freely choose n-1 values, then sum of the set of data can be written as [tex]\sum \sumx^{n-1} _i__=_1 Xi[/tex]

[tex]\sum \sumx^{n-1} _i__=_1 Xi \ =\ N \overline x[/tex]

The number of n values which is referred to the degree of freedom is n-1

Answer:

The recurrence relation for aₙ is  aₙ = 2aₙ - 1 + 2aₙ -2 ; is n≥ 3 with the initial conditions as a₁ =3; a₂ = 8

Step-by-step explanation:

Solution

Recurrence relation for n - digit ternary sequence with no occurrence of consecutive 0's in them.

Ternary sequence is sequence with each of digits either 0, 1 or 2.

Now

Let aₙ = denote the number of n - digit ternary sequence with no occurrence of consecutive 0's in them.

Let us first find few initial values of aₙ

For n = 1

a₁ represent the number of 1- digit ternary sequence with no occurrence of consecutive 0's in them.  

This 1-digit sequence can be either 0 or 1 or 2.

Thus,

a₁ = 3

For n =2

a₂ represent the number of 2- digit ternary sequence with no occurrence of consecutive 0's in them.

This 2-digit sequence can have either 0 or 1 or 2 as each of its two digit, but making sure that there are no two consecutive 0 in the sequence.

here are " 9 " 2-digit ternary sequence ........... (three choices for 1st digit and three choices for 2nd digit)  

But one of these 9 sequence there are consecutive 0's .... (00)  

So we eliminate this one sequence.

So, a₂ = 8

Now

let us find the recurrence relation

Fir n ≥ 3

aₙ s the number of n - digit ternary sequence with no occurrence of the consecutive 0's in them.

For the first case: if 1st digit of this n - digit ternary sequence is 1 or 2

Let assume the 1st digit of this n - digit ternary sequence is 1.  

Then for remaining n - 1 digits of this n - digit ternary sequence we have to make sure that there is no consecutive 0's in them.

For example, we have to form a n-1-digit ternary sequence with no occurrence of consecutive 0's in them which is by definition aₙ -1

So,

The number of n - digit ternary sequence with no occurrence of consecutive 0's in them if 1st digit of this sequence is 1 is aₙ -1.

Likewise, the number of n - digit ternary sequence with no occurrence of consecutive 0's in them if 1st digit of this sequence is 2 is aₙ -1.

So  

If 1st digit of this n - digit ternary sequence is 1 or 2, then the number of n - digit ternary sequence with no occurrence of consecutive 0's in them is shown as:

aₙ-1 + aₙ -1 = 2aₙ -1

For the second case: if 1st digit of this n - digit ternary sequence is 0

If 1st digit of this n - digit ternary sequence is 0, then the next digit cannot be 0 as well because that would make two consecutive 0's in the sequence Thus,

If 1st digit of this n - digit ternary sequence is 0, the next term can be either 1 or 2.

So there are 2 choices for 2nd digit.  

After this there are more n-2 digits.  

Then

For remaining n - 2 digits of this n - digit ternary sequence we have to make sure that there is no consecutive 0's in them  

For example, we have to form a n-2-digit ternary sequence with no occurrence of consecutive 0's in them. which is by definition aₙ - 2.

Now,

The total number of sequence in this case is given as:

2aₙ -2........... (2 choices for 2nd digit and aₙ - 2 choices for remaining n-2 digit)

Hence  

The number of n - digit ternary sequence with no occurrence of consecutive 0's in them if 1st digit of this sequence is 0 is aₙ = 2aₙ - 1 + 2aₙ -2 which is n≥ 3

Now,

The recurrence relation for aₙ is shown below:

aₙ = 2aₙ - 1 + 2aₙ -2; is n≥ 3

With the initial conditions as a₁ =3; a₂ = 8

Option D is the correct answer.

Answer:

D. ∠B and ∠C are congruent.

Step-by-step explanation:

Since, ∠A and ∠B are supplementary.

Therefore,

∠A + ∠B = 180°.....(1)

Since, ∠A and ∠C are supplementary.

Therefore,

∠A + ∠C = 180°.....(2)

From equations (1) & (2)

∠A + ∠B = ∠A + ∠C

=> ∠B = ∠C

Hence, ∠B and ∠C are congruent.

Answer:

[tex]y(x)=-0.028x+14.6[/tex]

Step-by-step explanation:

We are to write a linear equation that relates y in terms of x

The Birth Rate in 1994 =  14.6 births per thousand population.

The Birth Rate in 2004 =  14.32 births per thousand population.

A linear equation is of the form y=mx+b, where:

Step 1: Determine the birth rate per year

In 1994, x=0, y=14.6 thousands

In 2004, x=10, y=14.32 thousands

[tex]m=\dfrac{14.32-14.6}{10-0}\\=\dfrac{-0.28}{10}\\m=-0.028[/tex]

Substituting m into our linear equation, we have:

[tex]y(x)=-0.028x+b[/tex]

When x=10, y=14.32

[tex]14.32=-0.028(10)+b\\b=14.32+0.28\\b=14.6[/tex]

Therefore, a linear equation that relates y in terms of x is:

[tex]y(x)=-0.028x+14.6[/tex]

Answer:

Type of error: Run-on(comma splice).

Best revision to fix it: Adding a semicolon after beginners .

Explanation:

A run-on sentence is described as a sentence in which two independent clauses are joined inappropriately. It could be either comma splice where the two independent clauses are incorrectly linked using a comma or fused sentence when the two clauses run-on without employing appropriate coordinating conjunction or punctuation marks to separate the two ideas.

In the given sentence, it exemplifies a comma splice type of run-on sentence error. To fix this error, a semicolon after 'beginners' can be employed instead of a comma. This will help in connecting the two ideas appropriately where the first idea leads the second. Thus, the final sentence reads as:

'The guitar is another excellent instrument for beginners; however, it takes more practice than a recorder.'

Answer:

Many people play a musical instrument music can be soothing. A lot of schools teach the recorder; it is inexpensive and easy to play. The guitar is another excellent instrument for beginners, it takes more practice than a recorder.

What type of error is present in the underlined sentence?

✔ run-on

Which is the best revision to fix the error?

✔ adding a semicolon after instrument

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given the rate at which the volume V (in dollars) of sales is changing is approximated by the equation

V ' ( t ) = − 26400 e^− 0.49 t .

t = time (in days)

.v'(6) can be derived by simply substituting t = 6 into the modelled equation as shown:

V'(6) = − 26400 e− 0.49 (6)

V'(6) = -26400e-2.94

V'(6) = -26400×-0.2217

V'(6) = $5852.88

V'(6) = $5,853 to nearest dollars

V'(6) = 585300cents to nearest cent

To get v(6), we need to get v(t) first by integrating the given function as shown:

V(t) = ∫−26400 e− 0.49 t dt

V(t) = -26,400e-0.49t/-0.49

V(t) = 53,877.55e-0.49t + C.

When t = 0, V(t) = $170,000

170,000 = 53,877.55e-0 +C

170000 = 53,877.55(2.7183)+C

170,000 = 146,454.37+C

C = 170,000-146,454.37

C = 23545.64

V(6) = 53,877.55e-0.49(6)+ 23545.64

V(6) = -11,945.63+23545.64

V(6) = $11,600 (to the nearest dollars)

Since $1 = 100cents

$11,600 = 1,160,000cents

Answer:

19467649.76 lb-ft^2/s^2

Step-by-step explanation:

diameter of the pool d = 20 ft

radius = d/2 = 20/2 = 10 ft

height of pool side h = 6 ft

depth of water d = 5 ft

the force on the bottom of the pool due to the water in the pool is

F = pgdA

where p = density of water = 62.4 lb/ft^3

g = acceleration due to gravity = 32.17 ft/s^2

Area A = [tex]\pi r^{2}[/tex] = [tex]3.142 * 10^{2}[/tex] = 314.2 ft^2

Force on pool bottom = 64.2 x 32.17 x 5 x 314.2 = 3244608.29 lb-ft/s^2

work done = force times the height the water will be pumped

work = F x h = 3244608.29 x 6 = 19467649.76 lb-ft^2/s^2

The work (in ft-lb) is required to pump all of the water out over the side is :

Formula:

F = pgdA

p = density of water = 62.4 lb/ft^3

g = acceleration due to gravity = 32.17 ft/s^2

Area A = [tex]\pi r2\\[/tex] = 314.2 ft^2

The work (in ft-lb) is required to pump all of the water out over the side is 19467649.76 lb-ft^2/s^2.

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