4. A study of 30 fathers was completed in which the fathers were asked the highest level of education they had completed. 10 completed only elementary school, 10 completed elementary school and high school; 7 completed elementary school, high school and college; 3 completed elementary school, high school, college and graduate school. What was the cumulative percentage of fathers who completed only elementary school?

Answer:

Your Winnings are $ -1

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.

Since we are not told what two balls were actually drawn, then the expected winnings would be of the greatest probability occurring. To figure this out we need to calculate the probability of getting two red balls consecutively and the probability of getting two different colors consecutively. Since there are a total of 10 balls the probabilities of getting one ball would be the following,

Red: [tex]\frac{4}{10} =  40%[/tex]

Green: [tex]\frac{6}{10} = 60%[/tex]

Now we need to know the probability of us getting 2 reds together, two greens, or one of each.

2 Reds: [tex]\frac{4}{10} *\frac{4}{10} = \frac{16}{100}  = 16%[/tex]

2 Greens:  [tex]\frac{6}{10} *\frac{6}{10} = \frac{36}{100}  = 36%[/tex]

1 Each:  [tex]\frac{4}{10} *\frac{6}{10} = \frac{24}{100}  = 24%[/tex]

So we can see that the highest probability is getting 2 Greens in  a row. Meaning you would have to pay $1

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

If [tex]f(x)=4x-3[/tex]:

[tex]\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4[/tex]

If [tex]f(x)=4x^{-3}[/tex]:

[tex]\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}[/tex]

[tex]\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}[/tex]

[tex]\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}[/tex]

Answer:

4

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given that a random sample of 90 companies taken in 2006 showed that 14 offered​ high-deductible health insurance plans to their workers. A separate random sample of 120 firms taken in 2007 showed that 30 offered​ high-deductible health insurance plans to their workers.

H0: p1=p2

Ha: p1 <p2

(Two tailed test at99%)

Difference 14.44 %

Chi-squared 5.883

DF  1

Significance level P = 0.0153

Since p >0.01, our alpha reject null hypothesis.

NO. Based on the sample​ results, you can not  conclude that there is a higher proportion of companies offering​ high-deductible health insurance plans to their workers in 2007 than in 2006

Answer:

1,084 tickets were sold that cost $10

1,291 tickets were sold that cost $20

767 tickets were sold that cost $30

Step-by-step explanation:

Let

x ----> the number of tickets  that cost $10 sold

y ----> the number of tickets  that cost $20 sold

z ----> the number of tickets  that cost $30 sold

we know that

x+y+z=3,142 -----> equation A

10x+20y+30z=59,670 ----> equation B

y=x+207 ----> equation C

substitute equation C in equation A and equation B

x+(x+207)+z=3,142 ----> 2x+z=2,935 ----> equation D

10x+20(x+207)+30z=59,670 ---> 30x+30z=55,530 ----> equation E

Solve the system of equations D and E by graphing

The solution is the intersection point both graphs

The solution is the point (1,084,767)

so

x=1,084, z=767

see the attached figure

Find the value of y

y=x+207 ----> y=1,084+207=1,291

therefore

1,084 tickets were sold that cost $10

1,291 tickets were sold that cost $20

767 tickets were sold that cost $30

Answer:

The product of the y-coordinates of the solutions is equal to 3

Step-by-step explanation:

we have

[tex]x^{2}+4y^{2}=40[/tex] -----> equation A

[tex]x+2y=8[/tex] ------> equation B

Solve by graphing

Remember that the solutions of the system of equations are the intersection point both graphs

using a graphing tool

The solutions are the points (2,3) and (6,1)

see the attached figure

The y-coordinates of the solutions are 3 and 1

therefore

The product of the y-coordinates of the solutions is equal to

(3)(1)=3

Answer: Assuming the function is [tex]g(x)=\frac{2x+1}{x-3}[/tex]:

The x-intercept is [tex](\frac{-1}{2},0)[/tex].

The y-intercept is [tex](0,\frac{-1}{3})[/tex].

The horizontal asymptote is [tex]y=2[/tex].

The vertical asymptote is [tex]x=3[/tex].

Step-by-step explanation:

I'm going to assume the function is: [tex]g(x)=\frac{2x+1}{x-3}[/tex] and not [tex]g(x)=2x+\frac{1}{x}-3[/tex].

So we are looking at [tex]g(x)=\frac{2x+1}{x-3}[/tex].

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

[tex]0=\frac{2x+1}{x-3}[/tex]

A fraction is only 0 when it's numerator is 0.  You are really just solving:

[tex]0=2x+1[/tex]

Subtract 1 on both sides:

[tex]-1=2x[/tex]

Divide both sides by 2:

[tex]\frac{-1}{2}=x[/tex]

The x-intercept is [tex](\frac{-1}{2},0)[/tex].

The y-intercept is when x is 0.

Replace x with 0.

[tex]g(0)=\frac{2(0)+1}{0-3}[/tex]

[tex]y=\frac{2(0)+1}{0-3}[/tex]  

[tex]y=\frac{0+1}{-3}[/tex]

[tex]y=\frac{1}{-3}[/tex]

[tex]y=-\frac{1}{3}[/tex].

The y-intercept is [tex](0,\frac{-1}{3})[/tex].

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is [tex]y=\frac{2}{1}[/tex].  After simplifying you could just say the horizontal asymptote is [tex]y=2[/tex].

Or!

I could do some division to make it more clear.  The way I'm going to do this certain division is rewriting the top in terms of (x-3).

[tex]y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}[/tex]

[tex]y=2+\frac{7}{x-3}[/tex]

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

Answer:

We are given that The dependent variable is production; that is, it is assumed that different levels of production result from a different number of employees.

Number of Assemblers(x)  One-Hour Production(y) (units)

2                                                           11

4                                                           18

1                                                            7

5                                                           29

3                                                           20

a. Draw a scatter diagram.

Solution : Refer the attached figure  

b. Based on the scatter diagram, does there appear to be any relationship between the number of assemblers and production? Explain.

Solution: The equation that shows the relationship between the number of assemblers and production is [tex]y=5.1x+1.7[/tex]

Where y is One-Hour Production (units)  and x is the Number of Assemblers

c.Compute the correlation coefficient.

Solution:

Formula of correlation coefficient:[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{[n \sum x^2 -(\sum x)^2][n \sum y^2 -(\sum y)^2]}[/tex]

            x        y         xy       [tex]x^2[/tex]   [tex]y^2[/tex]

           2        11        22              4                   121

           4        18       72               16                 324

           1          7         7                 1                    49

           5         29      145             25                  841

           3         20      60               9                    400            

Sum:   15     85       306           55                  1735

n=5

Substitute the values in the formula :

[tex]r=\frac{5(306)-(15)(85)}{[5 (55) -(15)^2][5 (1735) -(85)^2]}[/tex]

[tex]r=0.00351[/tex]

The correlation coefficient is 0.00351

Answer:

[tex]P(5)=0.238[/tex]

[tex]P(x\geq 6)=0.478[/tex]

[tex]P(x<4)=0.114[/tex]

Step-by-step explanation:

In this case we can calculate the probability using the binomial probability formula

[tex]P(X=x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}[/tex]

Where p is the probability of obtaining a "favorable outcome " x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly​ five

This is:

[tex]x=5,\ n=10,\ p=0.54.[/tex]

So:

[tex]P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}[/tex]

[tex]P(5)=0.238[/tex]

(b) at least​ six

This is: [tex]x\geq 6,\ n=10,\ p=0.54.[/tex]

[tex]P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)[/tex]

[tex]P(x\geq 6)=0.478[/tex]

(c) less than four

This is: [tex]x< 4,\ n=10,\ p=0.54.[/tex]

[tex]P(x<4)=P(3) + P(2)+P(1)+P(0)[/tex]

[tex]P(x<4)=0.114[/tex]

This question is based on the probability. Therefore, the required probabilities  are :  (a) [tex]P(5) = 0.238[/tex],  (b)[tex]P(x \geq 6) = 0.478[/tex]  and (c) [tex]P(x <4) = 0.114[/tex].

Given:

54​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults.

We have to find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

According to the question,

[tex]P(5) = 0.238\\P(x \geq 6) = 0.478\\P(x <4) = 0.114[/tex]

In this we have to calculate the probability using the binomial probability formula,

[tex]P(X=x) = \dfrac{n!}{x!(n-x)!} \times p^{x} \times (1-p)^{n-x}[/tex]

Where, p is the probability of obtaining a "favorable outcome ", x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly​ five  

x=5, n= 10, p = 0.54

[tex]P(X=5)= \dfrac{10!}{5!(10-5)!} \times 0.5^{x} \times (1-0.54)^{10-5}[/tex]

P(X=5) = 0.238

(b) at least​ six

[tex]x\geq 6, n=10, p=0.54\\P(x\geq 6) = P(6) + P(7) + P(8) + P(9)+P(10)\\P(x\geq 6) = 0.478[/tex]

(c) less than four

[tex]x< 6, n=10, p=0.54\\P(x< 4) = P(3) + P(2) + P(1) + P(0)\\P(x< 4) = 0.114[/tex]

Therefore, the answers are :  (a) [tex]P(5) = 0.238[/tex],  (b)[tex]P(x \geq 6) = 0.478[/tex]

and (c) [tex]P(x <4) = 0.114[/tex].

For more details, please refer this link:

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Answer:

Part A) [tex]w+35 > 1,050[/tex]

Part B) [tex]w > 1,015\ pounds[/tex]

Part C) The graph in the attached figure

Step-by-step explanation:

Part A) Model with an inequality.

Let

w -----> the limit of weight in an elevator

we know that

The inequality that represent this situation is

[tex]w+35 > 1,050[/tex]

Part B)   Solve the inequality

[tex]w+35 > 1,050[/tex]

solve for w

Subtract 35 sides

[tex]w+35-35 > 1,050-35[/tex]

[tex]w > 1,015\ pounds[/tex]

All real numbers greater than 1,015 pounds

Part C) Where would the solution for the inequality be graphed on the number line?

[tex]w > 1,015\ pounds[/tex]

The solution is the interval -------> (1,015,∞)

In a number line is the shaded area at right of the number x=1.015 (open circle)

see the attached figure

Answer:  [tex]\dfrac{21}{31}[/tex]

Step-by-step explanation:

Let A represents the event that high school dropouts are ​16- to​ 17-year-olds and B be the event that high school dropouts are ​white.

Given : The probability of high school dropouts are​ 16- to​ 17-year-olds :[tex]P(A)=0.093[/tex]

The probability of of high school dropouts are white​ 16- to​ 17-year-olds :

[tex]P(A\cap B)=0.063[/tex]

Then , the conditional  probability that a randomly selected dropout is​ white, given that he or she is 16 to 17 years​ old is given by :-

[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{0.063}{0.093}=\dfrac{21}{31}[/tex]

Answer: 0.8351

Step-by-step explanation:

Given :Mean : [tex]\mu=\text{ 3 inches}[/tex]

Standard deviation : [tex]\sigma =\text{ 0.5 inch}[/tex]

The formula for z -score :

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

For x= 2.5 ,

[tex]z=\dfrac{2.5-3}{0.5}=-1[/tex]

For x= 4.25 ,

[tex]z=\dfrac{4.25-3}{0.5}=2.5[/tex]

The p-value = [tex]P(-1<z<2.5)=P(z<2.5)-P(z<-1)[/tex]

[tex]=0.9937903-0.1586553=0.835135\approx0.8351[/tex]

Hence,  the probability of picking an apple with diameter between 2.5 and 4.25 inches =0.8351.

To find the probability, we standardize the values, convert them into z-scores, look up the z-scores in a z-table, and subtract the lower cumulative probability from the higher one. The probability of picking an apple with diameter between 2.5 inches and 4.25 inches is 83.51%.

The problem involves a normal distribution where the diameter of Red Delicious apples has a mean of 3 inches and a standard deviation of 0.5 inch. The objective is to find the probability of picking an apple with a diameter between 2.5 and 4.25 inches.

We first standardize the given values to convert them into z-scores by subtracting the mean from the given value and dividing by the standard deviation. For 2.5 inches, z = (2.5 - 3) / 0.5 = -1. For 4.25 inches, z = (4.25 - 3) / 0.5 = 2.5.

Using a z-table, the z-score of -1 corresponds to a cumulative probability of 0.1587 and the z-score of 2.5 corresponds to a cumulative probability of 0.9938. To find the probability between these two diameters, we subtract the cumulative probability of -1 from the cumulative probability of 2.5.

Therefore, the probability of picking an apple with diameter between 2.5 inches and 4.25 inches is 0.9938 - 0.1587 = 0.8351 or 83.51%.

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Answer:

a) 40, 50, 55, 57.5

b) [tex]S_n=h_0+2h_0\sum_{n=1}^{\infty}r^n[/tex]

Step-by-step explanation:

h₀ = Initial height of the ball =20

r = Rebound fraction = 0.5

a) The series of bouncing balls is given by

Sₙ=h₀+2h₀(r¹+r²+r³+r⁴.........rⁿ)

S₁=h₀+2h₀r¹=20+2×20×0.5=40

S₂=h₀+2h₀(r¹+r²)=20+2×20×(0.5+0.5²)=50

S₃=h₀+2h₀(r¹+r²+r³)=20+2×20×(0.5+0.5²+0.5³)=55

S₄=h₀+2h₀(r¹+r²+r³+r⁴)=20+2×20×(0.5+0.5²+0.5³+0.5⁴)=57.5

b) General expression for the nth term of the sequence

[tex]S_n=h_0+2h_0\sum_{n=1}^{\infty}r^n[/tex]

Final answer:

The first 4 heights after each bounce form a geometric sequence with the first term h₀ being 20 meters and subsequent terms being 10, 5, and 2.5 meters. The general expression for the nth term (hn) is given by the formula hₙ = 20 * (0.5)ⁿ

Explanation:

Given an initial height h₀ of 20 meters and a rebound fraction r of 0.5, the sequence of heights after each bounce forms a geometric sequence.

The first height is h₀ which is 20 meters. Subsequent heights can be found by multiplying the previous height by the rebound fraction r.

Third term : h2 = h= 10 * 0.5 = 5 meters

General Expression for the nth Term (b)

The nth term (hₙ) of the sequence can be found using the formula for the nth term of a geometric sequence:

hₙ = h₀ * rⁿ

For this particular sequence:

hₙ = 20 * (0.5)ⁿ

Answer:

b) 4,000,000

Step-by-step explanation:

Let the population is measured since 1995,

Given,

The initial population, P = 3,942,000,

Annual rate of growing, r = 0.7% = 0.007,

If y represents the population after t years

So, the population after t years would be,

[tex]y=Pe^{rt}[/tex]

[tex]y=3942000(2.7)^{0.007x}[/tex]

Therefore, the population after 5 years,

[tex]y=3942000(2.7)^{0.007\times 5}=3942000(2.7)^{0.035}=4081448.78924\approx 4000000[/tex]

Hence, the population in 2000 would be approximately 40,00,000.

Option 'b' is correct.

By using the accurate population growth formula P = P0(1 + r)^n and substitifying the given values, the population of the Asian country in 2000 would be approximately 4,160,000 residents.

The question involves the concept of exponential growth, specifically applied to the population growth of an Asian country. Now, in given formula y = 3,942,000(2.7)^0.0074, we should use the correct growth rate formula to solve the problem which should be P = P0(1 + r)^n since the population growth is annually and continuous. Here, P0 is the initial population (3,942,000), r is the growth rate (0.7% or 0.007) and n is the number of years (2000-1995 = 5 years).

Using this formula, if you substitute these values in, you should get:

P = 3,942,000(1 + 0.007)^5

If you calculate this out, you reach a population of approximately 4,160,000 residents at the end of year 2000. Thus, answer 'c' is the correct option. Please remember exponential growth is a concept necessary for understanding population dynamics across multiple fields like demography, biology and mathematical modeling.

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Answer:

  (d)  10

Step-by-step explanation:

Multiply by x and divide by its coefficient:

  (43.65)(8.79) = (0.4365)(87.9)x

  (43.65)(8.79)/((0.4365)(87.9)) = x

At this point, any calculator can give you the answer. It is, perhaps, more satisfying to work out the answer without a calculator.

  x = (43.65)/(0.4365) × (8.79)/(87.9)

In the first quotient, the numerator is 100 times the denominator; in the second, the denominator is 10 times the numerator.

  x = (100) × (1/10) = 100/10

  x = 10

_____

Moving the decimal point to the right 1 place multiplies the numerical value by 10.

The value of x in the given equation that satisfies the condition is 10.

In this question, we're given a mathematical expression in which the value of x is unknown. We're looking for the value of x that satisfies the equation:

( 43.65 ) ( 8.79 ) / x = ( 0.4365 ) ( 87.9 )

To solve this equation for x, we can start by noting the similarity between the left and right sides. We have larger numbers on the left side that appear, in reduced form, on the right side.

Follow these steps:

So, the correct option would be (d) 10.

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The question is about calculating the conditional probability density functions of two random variables X and Y given their joint density function. However, without the marginal density functions for X and Y, it is not possible to provide exact numerical answers. General formulas involve dividing the joint density function by the respective marginal density function.

The question relates to the field of probability and statistics, specifically pertaining to joint density functions of two random variables (X and Y). We have the joint density function f(x, y) = 6xe^(−x(y+6)) where x > 0 and y > 0.

For the conditional probability density function, the general formula for fX|Y(x, y) is the joint density function divided by the marginal density function of Y. Similarly, for fY|X(x, y), it is the joint density function divided by the marginal density function of X.

However, the marginal density functions for X and Y are not specified in the question. Typically, to find these, one would integrate the joint density function with respect to the other variable. Due to this missing information, it's not possible to provide an exact numerical answer.

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Answer:

40

Step-by-step explanation:

8: 8, 16, 24, 32, 40

10: 10 20 30 40

Answer:

The least common multiple of 8 and 10 is 40.

Step-by-step explanation:

8: 8*1=8, 8*2=16, 8*3=24, 8*4=32, 8*5=40

10: 10*1=10, 10*2=20, 10*3=30, 10*4=40

The least common factor of 8 and 10 is 40.

Answer: 0.30

Step-by-step explanation:

Binomial distribution formula :-

[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success in x trials , n is the total number of trials and p is the probability of getting success in each trial.

Given : The probability that a shipment are known to be defective= 0.07

If a sample of 5 items is randomly selected from this shipment,then the probability that at least one defective item will be observed in this sample will be :-

[tex]P(X\geq1)=1-P(0)\\\\=1-(^5C_0(0.07)^0(1-0.07)^{5-0})\\\\=1-(0.93)^5=0.3043116307\approx0.30[/tex]

Hence, the probability that at least one defective item will be observed in this sample =0.30

Answer: 0.9834

Step-by-step explanation:

Given : The test scores are normally distributed with

Mean : [tex]\mu=\ 0[/tex]

Standard deviation :[tex]\sigma= 1[/tex]

The formula to calculate the z-score :-

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

For x = -2.13

[tex]z=\dfrac{-2.13-0}{1}=-2.13[/tex]

For x = 3.88

[tex]z=\dfrac{3.88-0}{1}=3.88[/tex]

The p-value = [tex]P(-2.13<z<3.88)=P(z<3.88)-P(z<-2.13)[/tex]

[tex]0.9999477-0.0165858=0.9833619\approx0.9834[/tex]

Hence, the probability that a given score is between negative 2.13 and 3.88 = 0.9834

Answer:

a) The value of the investment is $33176.506

b) The value of the investment is $33555.53

c) The value of the investment is $33893.36

d) The value of the investment is $33961.33

Step-by-step explanation:

This is a compound interest problem

Compound interest formula:

The compound interest formula is given by:

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

A: Amount of money(Balance)

P: Principal(Initial deposit)

r: interest rate(as a decimal value)

n: number of times that interest is compounded per unit t

t: time the money is invested or borrowed for.

In our problem, we have:

A: the value we want to find

P = 20600(the value invested)

r = 0.1

n: Will change for each letter

t = 5.

a) If the interest is compounded anually, n = 1. So.

[tex]A = 20600(1 + \frac{0.1}{1})^{1*5}[/tex]

[tex]A = 33176.506[/tex]

The value of the investment is $33176.506

b) If the interest is compounded semianually, it happens twice a year, which means n = 2. So:

[tex]A = 20600(1 + \frac{0.1}{2})^{2*5}[/tex]

[tex]A = 33555.23[/tex]

The value of the investment is $33555.53

c) If the interest is compounded monthly, it happens 12 times a year, which means n = 12. So:

[tex]A = 20600(1 + \frac{0.1}{12})^{12*5}[/tex]

[tex]A = 33893.36[/tex]

The value of the investment is $33893.36

d) If the interest is compounded monthly, it happens 365 times a year, which means n = 365. So:

[tex]A = 20600(1 + \frac{0.1}{365})^{365*5}[/tex]

[tex]A = 33961.33[/tex]

The value of the investment is $33961.33

Final answer:

The future value of $20,600 invested at a 10% annual interest rate after 5 years varies based on the compounding frequency: annually it will be $33,186.35, semiannually $33,644.31, monthly $33,949.69, and daily $34,030.18.

Explanation:

Calculating Future Value with Different Compounding Methods

To calculate the future value of an investment with compound interest, we use the formula  [tex]FV =PV(1+r/n)^{nt}[/tex],where FV is the future value, P is the principal amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is time in years.

Given a principal amount of $20,600 and an annual interest rate of 10%, we will calculate the value of the investment at the end of 5 years for the following compounding methods:

Annual compounding (n=1): FV = 20600(1 + 0.10/1)⁽¹ˣ⁵⁾

Semiannual compounding (n=2): FV = 20600(1 + 0.10/2)⁽²ˣ⁵⁾

Monthly compounding (n=12): FV = 20600(1 + 0.10/12)⁽¹²ˣ⁵⁾

Daily compounding (n=365): FV = 20600(1 + 0.10/365)⁽³⁶⁶ˣ⁵⁾

Using a calculator or spreadsheet software, we can find the values to the nearest cent:

(a) Annual: $33,186.35

(b) Semiannual: $33,644.31

(c) Monthly: $33,949.69

(d) Daily: $34,030.18

The compounding effect shows that the more frequently the interest is compounded, the greater the final value of the investment.

Answer with explanation:

a. f(x)=3x-4

Let [tex]f(x_1)=f(x_2)[/tex]

[tex]3x_1-4=3x_2-4[/tex]

[tex]3x_1=3x_2-4+4[/tex]

[tex]3x_1=3x_2[/tex]

[tex]x_1=x_2[/tex]

Hence, the function one-one.

Let f(x)=y  

[tex]y=3x-4[/tex]

[tex]3x=y+4[/tex]

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

We can find pre image in domain R for every y in range R.

Hence, the function onto.

b.g(x)=[tex]x^2-2[/tex]

Substiute x=1

Then [tex]g(x)=1-2=-1[/tex]

Substitute x=-1

Then g(x)=1-2=-1

Hence, the image of 1 and -1 are same . Therefore, the given function g(x) is not one-one.

The given function g(x) is not onto because there is no pre image of -2, -3,-4......  R.

Hence, the function neither one-one nor onto on given  R.

c.[tex]h(x)=\frac{2}{x}[/tex]

The function is not defined for x=0 .Therefore , it is not a function on domain R.

Let [tex]h(x_1)=h(x_2)[/tex]

[tex] \frac{2}{x_1}=\frac{2}{x_2}[/tex]

By cross mulitiply

[tex]x_1= \frac{2\times x_2}{2}[/tex]

[tex]x_1=x_2[/tex]

Hence, h(x) is a one-one function on R-{0}.

We can find pre image for every value of y except zero .Hence, the function

h(x) is onto on R-{0}.

Therefore, the given function h(x) is both one- one and onto on R-{0} but not on R.

d.k(x)= ln(x)

We know that logarithmic function not defined for negative values of x. Therefore, logarithmic is not a function R.Hence, the given function K(x) is not a function on R.But it is define for positive R.

Let[tex]k(x_1)=k(x_2)[/tex]

[tex] ln(x_1)=ln(x_2)[/tex]

Cancel both side log then

[tex]x_1=x_2[/tex]

Hence, the given function one- one on positive R.

We can find pre image in positive R for every value of [tex]y\in R^+[/tex].

Therefore, the function k(x) is one-one and onto on [tex]R^+[/tex] but not on R.

e.l(x)=[tex]e^x[/tex]

Using horizontal line test if we draw a line y=-1 then it does not cut the graph at any point .If the horizontal line cut the graph atmost one point the function is one-one.Hence, the horizontal line does not cut the graph at any point .Therefore, the function is one-one on R.

If a horizontal line cut the graph atleast one point then the function is onto on a given domain and codomain.

If we draw a horizontal line y=-1 then it does not cut the graph at any point .Therefore, the given function is not onto on R.

The correct answers are:

(a) [tex]P(0\leq Z\leq 2.38)=0.4913[/tex]

(b)  [tex]P(0\leq Z\leq 1)=0.3413[/tex]

(c)  [tex]P(-2.70\leq Z\leq 0)=0.4965[/tex]

(d)  [tex]P(-2.70\leq Z\leq 2.70)=0.9926[/tex]

(e)  [tex]P(Z\leq 1.62)=0.9474[/tex]

(f)  [tex]P(-1.55\leq Z)=0.9394[/tex]

(g)  [tex]P(-1.70\leq Z\leq 2.00)=0.9326[/tex]

(h)  [tex]P(1.62\leq Z\leq 2.50)=0.0457[/tex]

(i) [tex]P(1.70\leq Z)=0.0446[/tex]

(j) [tex]P(|Z|\leq 2.50)=1.9862[/tex]

Let's calculate these probabilities step by step using the standard normal distribution table (also known as the z-table).

For reference, the standard normal distribution table provides the probabilities associated with the standard normal random variable (Z), which has a mean of [tex]0[/tex] and a standard deviation of [tex]1[/tex].

We'll use the standard normal distribution table to find the probabilities corresponding to the given Z-values.

(a [tex]P(0\leq Z\leq 2.38)[/tex] From the z-table,  [tex]P(Z\leq 2.38)=0.9913 P(0\leq Z\leq 2.38)=0.9913-0.5=0.4913[/tex] (subtracting the cumulative probability up to [tex]0[/tex] from the cumulative probability up to [tex]2.38[/tex])

(b) [tex]P(0\leq Z\leq 1)[/tex] From the z-table,  [tex]P(Z\leq 1)=0.8413 P(0\leq Z\leq 1)=0.8413-0.5=0.3413[/tex]

(c) [tex]P(-2.70\leq Z\leq 0)[/tex] From the z-table,  [tex]P(Z\leq 0)=0.5[/tex] and  [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 0)=0.5-0.0035=0.4965[/tex]

(d)  [tex]P(-2.70\leq Z\leq 2.70)[/tex] From the z-table,  [tex]P(Z\leq 2.70)=0.9961[/tex] and  [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 2.70)=0.9961-0.0035=0.9926[/tex]

(e)  [tex]P(Z\leq 1.62)[/tex] From the z-table,  [tex]P(Z\leq 1.62)=0.9474[/tex]

(f)  [tex]P(-1.55\leq Z)[/tex] From the z-table,  [tex]P(Z\leq -1.55)=0.0606 P(-1.55\leq Z)=1-0.0606=0.9394[/tex] (subtracting the cumulative probability up to [tex]-1.55[/tex] from [tex]1[/tex])

(g [tex]P(-1.70\leq Z\leq 2.00)[/tex] From the z-table,  [tex]P(Z\leq 2.00)=0.9772[/tex] and  [tex]P(Z\leq -1.70)=0.0446 P(-1.70\leq Z\leq 2.00)=0.9772-0.0446=0.9326[/tex]

(h)   [tex]P(1.62\leq Z\leq 2.50)[/tex] From the z-table,   [tex]P(Z2.50)=0.9931[/tex] and  [tex]P(Z\leq 1.62)=0.9474 P(1.62\leq Z\leq 2.50)=0.9931-0.9474=0.0457[/tex]

(i) [tex]P(1.70\leq Z)[/tex][tex]From the z-table, P(Z\leq 1.70)=0.9554 , P(1.70\leq Z)=1-0.9554=0.0446[/tex] (subtracting the cumulative probability up to [tex]1.70[/tex] from [tex]1[/tex])

(j) [tex]P(|Z|\leq 2.50)[/tex] Since the standard normal distribution is symmetric,   [tex]P(|Z|\leq 2.50)=2*P(0\leq Z\leq 2.50)=2*0.9931=1.9862[/tex] (multiply by [tex]2[/tex] because the probability of Z being between [tex]-2.50[/tex] and [tex]2.50[/tex] is twice the probability of Z being between [tex]0[/tex] and [tex]2.50[/tex] )

Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.)

(a) [tex]P(0 \leq Z \leq 2.38) .4913[/tex]

(b) [tex]P(0 \leq Z\leq 1) .3413[/tex]

(c) [tex]P(-2.70 \leq Z \leq 0) .4965[/tex]

(d) [tex]P(-2.70 \leq Z \leq 2.70) .9931[/tex]

(e) [tex]P(Z \leq 1.62) .9474[/tex]

(f) [tex]P(-1.55 \leq Z) .9394[/tex]

(g) [tex]P(-1.70 \leq Z \leq 2.00) .9327[/tex]

(h) [tex]P(1.62 \leq Z \leq 2.50) .0464[/tex]

(i) [tex]P(1.70 \leq Z) .0445[/tex]

(j) [tex]P(|Z| \leq 2.50) .9876[/tex]

Answer with explanation:

The given non Homogeneous linear differential equation is:

   y" +4 y'=3 Sin 2 x-------(1)

Put , u=y'

Differentiating once

u'=y"

Substituting the value of , y' and y" in equation (1)

⇒u' +4u =3 Sin 2x

This is a type of linear differential equation.

Integrating factor [tex]=e^{4t}[/tex]

Multiplying both sides of equation by Integrating factor

[tex]e^{4 x}(u'+4u)=e^{4x}3 \sin 2x\\\\ \text{Integrating both sides}\\\\ue^{4x}=\int {3 \sin 2x \times e^{4x}} \, dx \\\\ue^{4x}=\frac{3e^{4x}}{2^2+4^2}\times (4\sin 2x -2 \cos 2x)\\\\ue^{4x}=\frac{3e^{4x}}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}\\\\ \text{Using the formula of}\\\\\int{e^{ax}\sin bx } \, dx=\frac{e^{ax}}{a^2+b^2}\times (a \sin bx-b \cos bx)+C[/tex]

where C and [tex]C_{1}[/tex] are constant of integration.

Replacing , u by , y' in above equation we get the solution of above non homogeneous differential equation

  [tex]y'(x)=\frac{3}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}e^{-4 x}[/tex]

Answer:

In year 2013 annual maintenance done = $ 3,000,000

Out of this 85% is expected to renew in 2014 = [tex]0.85\times3000000[/tex] = $2,550,000

Now Sales in 2013 = $3,000,000

60% of these sales = [tex]0.60\times3000000[/tex] = $1,800,000

Now out of this 60%, 20% annual maintenance was expected to be paid

In 2014 = [tex]0.20\times1800000[/tex] = $360000

So, Total annual maintenance in 2014 = [tex]2550000+360000[/tex]

= $2,910,000

Answer:

1) [tex]\frac{7}{20}[/tex]

2) [tex]1\frac{1}{12}[/tex]

3) [tex]\frac{7}{8}[/tex]

Step-by-step explanation:

1) Given problem,

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

LCM(5,4) = 20,

[tex]\frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20}[/tex]

[tex]\frac{1}{4}=\frac{1\times 5}{4\times 5}=\frac{5}{20}[/tex]

[tex]\implies \frac{3}{5}-\frac{1}{4}=\frac{12}{20}-\frac{5}{20}=\frac{7}{20}[/tex]

2) Given problem,

[tex]\frac{5}{6}+\frac{1}{4}[/tex]

LCM(6, 4) = 12,

[tex]\frac{5}{6}=\frac{5\times 2}{6\times 2}=\frac{10}{12}[/tex]

[tex]\frac{1}{4}=\frac{1\times 3}{4\times 3}=\frac{3}{12}[/tex]

[tex]\implies \frac{5}{6}+\frac{1}{4}=\frac{10}{12}+\frac{3}{12}=\frac{13}{12}=1\frac{1}{12}[/tex]

3) ∵ Pizza ate by Manny = [tex]\frac{1}{4}[/tex]

Pizza ate by Frank = [tex]\farc{5}{8}[/tex]

∴ Total pizza eaten = [tex]\frac{1}{4}+\frac{5}{8}[/tex]

LCM(4, 8) = 8,

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

Thus, total pizza eaten =  [tex]\frac{2}{8}+\frac{5}{8}=\frac{7}{8}[/tex]

Answer:

[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]

Or

[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]

Step-by-step explanation:

We want to compute the curvature of the parameterized curve, [tex]r(t)=\:<\:6+5t^2,t,0)\:>\:[/tex] using the alternative formula:

[tex]k=\frac{|a\times v|}{|v|^3}[/tex].

We first compute the required ingredients.

The velocity vector is [tex]v=r'(t)=<\:10t,1,0\:>[/tex]

The acceleration vector is given by [tex]a=r''(t)=<\:10,0,0\:>[/tex]

The magnitude of the velocity vector is [tex]|v|=\sqrt{(10t)^2+1^2+0^2}=\sqrt{100t^2+1}[/tex]

The cross product of the velocity vector and the acceleration vector is

[tex]a\times v=\left|\begin{array}{ccc}i&j&k\\10&0&0\\10t&1&0\end{array}\right|=10k[/tex]

We now substitute ingredients into the formula to get:

[tex]k=\frac{|10k|}{(\sqrt{100t^2+1})^3}[/tex].

[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]

Or

[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]

Answer:

The correct option is No and lurking variable is "Economy".

Step-by-step explanation:

Consider the provided information.

It is given that the linear correlation between violent crime rate and percentage of the population that has a cell phone is minus 0.918 for years since 1995.

There is a lurking variable between the proportion of the cell phone population and the rate of violent crime, the correlation between the two factors does not indicate causation.

Once the economy become stronger, crime rate tends to decrease.

Thus the correct option is NO.

Now, we need to identify the lurking variable between percentage of the population with a cell phone and violent crime​ rate.

Lurking variable is the variable which is unknown and not controlled.

Here, if we observe we can identify that economy plays a vital role. If economy become stronger, crime rate tends to decrease and population are better to buy phones. Also it is difficult to controlled over it.

Thus, the lurking variable is "Economy".

Answer:

As given in the problem statement

mean score was less than 106

H0:mean score<106

so There is sufficient evidence to reject the claim μ < 106. is correct option.

Answer:

The number of cases prior to the increase is 50.

Step-by-step explanation:

It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.

We need to find the number of cases prior to the increase.

Let x be the number of cases prior to the increase.

x + 13.6% of x = 57

[tex]x+x\times \frac{13.6}{100}=57[/tex]

[tex]x+0.136x=57[/tex]

[tex]1.136x=57[/tex]

Divide both the sides by 1.136.

[tex]\frac{1.136x}{1.136}=\frac{57}{1.136}[/tex]

[tex]x=50.176[/tex]

[tex]x\approx 50[/tex]

Therefore the number of cases prior to the increase is 50.

Answer:

Step-by-step explanation:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30, 33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

a) Smoothing by bin means

Each bin has depth of 3

Dividing data into bins

so, Bin 1= 13, 15, 16

Bin 2=  16, 19, 20

Bin 3=  20, 21, 22

Bin 4= 22, 25, 25

Bin 5= 25, 25, 30

Bin 6= 33, 33, 35

Bin 7 = 35, 35, 35

Bin 8= 36, 40, 45

Bin 9=  46, 52, 70

Now, smoothing data by bin mean

so, Bin 1= 13, 15, 16 = (13+15+16)/3 = 15 Bin 1 = 15,15,15

Bin 2=  16, 19, 20 = (16+19+20)/3 = 18 Bin 2 = 18,18,18

Bin 3=  20, 21, 22= (20+21+22)/3 = 21 Bin 3 = 21,21,21

Bin 4= 22, 25, 25 = (22+25+25)/3 = 24 Bin 4 = 24,24,24

Bin 5= 25, 25, 30 =(25+25+30)/3 = 27 Bin 5 = 27,27,27

Bin 6= 33, 33, 35 = (33+33+35)/3 = 34 Bin 6 = 34,34,34

Bin 7 = 35, 35, 35 = (35+35+35)/3 = 35 Bin 7 = 35,35,35

Bin 8= 36, 40, 45 = (36+40+45)/3 = 40 Bin 8 = 40,40,40

Bin 9=  46, 52, 70=(46+52+70)/3 = 56 Bin 9 = 56,56,56

This technique is used to smooth the data. Data may have noise, using binning techniques we can remove noise from the data. It helps in providing more accurate results

b) How might you determine outliers in the data?

Outliers are the data that are abnormal to other data points. Outliers can be found by Box and whisker chart (box plot). Inter Quartile range can also be used to identify outliers

c)  What other methods are there for data smoothing?

Other methods of smoothing data are

a) binning by boundaries

b) Exponential smoothing

c) Random walk

The correct statement will be that the method of binning is extensively used for smoothening the data and cancels the noise a data contains. Outliers of a data can be determined using the quartile range.

There are different methods of data smoothening, such as binning with the help of boundaries, binning with the help of outliers, exponential binning, etc.

The bins will be created using the depth of 3 as,

Bin 1- 13,15,16 ; Bin 2- 16,19,20 ; Bin 3 20, 21, 22 ; Bin 4- 22,25,25 ; Bin 5- 25, 25, 30 ; Bin 6- 33,33,35 ; Bin 7- 35,35,35 ; Bin 8- 36,40,45 ; Bin 9- 46,52,70.

The vales of bins will be averaged and rounded off to the nearest whole numbers, considering bins as b in the following way.

[tex]b1= \dfrac {13+15+16}{3}\\\\b1= 14.67[/tex]

Continuing further in similar ways, we will find the values of remaining bins as

The outliers in the data can be found by using the box plot method and quartile range functions. Normally, such outliers are referred to as the mismatching data in a bin.

There are various different methods for smoothing the data. Some ways of smoothing the data are exponential smoothing, random walk smoothing, boundary binning method, etc.

Hence, the data is smoothened using the binning method and the values obtained are as above.

Learn more about bin data here:

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