A loan of $1000 is to be paid back, with interest, at the end of 1 year. Aft er 3 months, a partial payment of $300 is made. Use the US Rule to determine the balance due at the end of one year, considering the partial payment. Assume a simple interest rate of 9%.

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:

The closest conversion would be C. 30 ml equals 1 fluid ounce , it is only off by 0.43 ml

Step-by-step explanation:

Great question, it is always good to ask away in order to get rid of any doubts you may be having.

The metric system is a decimal system of measurement while the household system is a system of measurement usually found with kitchen utensils. The correct conversions are the following.

5 ml equals 0.33814 tablespoon

15 ml equals 3.04326 teaspoon

29.5735 ml equals 1 fluid ounce

236.588 ml equals 1 measuring cup

So the closest conversion would be C. 30 ml equals 1 fluid ounce , it is only off by 0.43 ml

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

The correct conversion between the metric and household system provided in the choices is 30ml equals 1 fluid ounce. However, 5ml is equivalent to 1 teaspoon, 15 ml to 1 tablespoon, and 250 ml to 1 measuring cup.

The correct conversion from the metric system to the household system among the options given is C. 30 ml equals 1 fluid ounce. The rationale behind this is that 30 ml is universally accepted as being equal to 1 fluid ounce in the household system.

Option A, B and, D are incorrect conversions. More accurate conversions would be: A. 5 ml equals 1 teaspoon; B. 15 ml equals 1 tablespoon; D. 250 ml equals 1 measuring cup.

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Answer:

  C).  sign chart; zeroes

Step-by-step explanation:

A function potentially changes sign at each of its zeros and vertical asymptotes. So, to fill out a sign chart, you need to determine what the sign is on either side of each of these points. You can do that using test numbers, or you can do it by understanding the nature of the zero or asymptote.

Examples:

f1(x) = (x -3) . . . . changes sign at the zero x=3. Is positive for x > 3, negative for x < 3.

f2(x) = (x -4)^2 . . . . does not change sign at the zero x=4. It is positive for any x ≠ 4. This will be true for any even-degree binomial factor.

f3(x) = 1/(x+2) . . . . has a vertical asymptote at x=-2. It changes sign there because the denominator changes sign there.

f4(x) = 1/(x+3)^2 . . . . has a vertical asymptote at x=-3. It does not change sign there because the denominator is of even degree and does not change sign there.

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 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.

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.)

a)

The probability that both televisions work is:  0.42

b)

The probability at least one of the two televisions does not​ work is:

                          0.5833

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

          9-3=6

a)

The probability that both televisions work is calculated by:

[tex]=\dfrac{6_C_2}{9_C_2}[/tex]

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

[tex]=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42[/tex]

b)

The probability at least one of the two televisions does not​ work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

[tex]=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5[/tex]

and the probability both do not work is:

[tex]=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833[/tex]

Hence, Probability that atleast does not work is:

             0.5+0.0833=0.5833

To find the probability that both televisions work, use the combination formula to determine the number of ways to select 2 working televisions out of the total number of televisions. Divide this number by the total number of ways to select 2 televisions.

To find the probability that both televisions work, we need to first determine the number of ways we can select 2 televisions out of the 9 available. This can be done using the combination formula, which is C(n, r) = n!/(r!(n-r)!), where n is the total number of items and r is the number of items being selected. In this case, n = 9 and r = 2.

Now we need to determine the number of ways we can select 2 working televisions out of the 6 working televisions. Again, we can use the combination formula with n = 6 and r = 2.

The final step is to divide the number of ways we can select 2 working televisions by the total number of ways we can select 2 televisions to get the probability that both televisions work.

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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:

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:

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:

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]\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:

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 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:  The correct option is (1) Dependent.

Step-by-step explanation:  For two events, we are given the following values of the probabilities :

P(A ∩ B) = 0.20,   P(A) = 0.49   and    P(B) = 0.41.

We are to check whether the events A and B are independent or dependent.

We know that

the two events C and D are said to be independent if the probabilities of their intersection is equal to the product of their probabilities.

That is,  P(C ∩ D) = P(C) × P(D).

For the given two events A and B, we have

[tex]P(A)\times P(B)=0.49\times0.41=0.2009\neq P(A\cap B)=0.20\\\\\Rightarrow P(A\cap B)\neq P(A)\times P(B).[/tex]

Therefore, the probabilities of the intersection of two events A and B is NOT equal to the product of the probabilities of the two events.

Thus, the events A and B are NOT independent. They are dependent events.

Option (1) is CORRECT.

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:

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:

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: 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:

[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:

4

Step-by-step explanation:

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

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:

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:Find the slope of the line that passes through the points shown in the table.

The slope of the line that passes through the points in the table is

.

Step-by-step explanation:

By substitifying the given points into the objective function, we can evaluate the minimum P. The point (x, y) = (0, 0) gives the minimum value of P = 0, which is the optimal solution for this problem.

This problem is a classic example of a linear programming problem, a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. In this case, we are asked to minimize P = 3x + 15y subject to the constraints [tex]2x + 4y \leq 12, 5x + 2y \leq 10, and ,x \geq 0, y \geq 0.[/tex] In other words, we are looking for values of x and y that satisfy the constraints and result in the smallest possible value of P.

By substituting our given points into the equation for P we can compare the results. The smallest value for P corresponds to the point (x, y) = (0, 0) with P = 0. This is the optimal solution for this problem because it results in the lowest value for P while still satisfying all the constraints.

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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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Answer: first option.

Step-by-step explanation:

By definition, the measure of any interior angle of an equilateral triangle is 60 degrees.

Then, we can find the value of "y". This is:

[tex]2y+6=60\\\\y=\frac{54}{2}\\\\y=27[/tex]

Since the three sides of an equilateral triangle have the same length, we can find the value of "x". This is:

[tex]x+4=2x-3\\\\4+3=2x-x\\\\x=7[/tex]

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:

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 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]