Marla Opper currently earns $95,000 a year and is offered a job in another city for $118,000. The city she would move to has 11 percent higher living expenses than her current city.



What amount must Marla earn in the new city to maintain her current buying power?

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: 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:

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]

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: The solution is,

[tex]x_1\approx 0.876[/tex]

[tex]x_2\approx 0.419[/tex]

[tex]x_3\approx 0.574[/tex]

Step-by-step explanation:

Given equations are,

[tex]8x_1 + x_2 + x_3 = 8[/tex]

[tex]2x_1 + 4x_2 + x_3 = 4[/tex]

[tex]x_1 + 3x_2 + 5x_3 = 5[/tex],

From the above equations,

[tex]x_1=\frac{1}{8}(8-x_2-x_3)[/tex]

[tex]x_2=\frac{1}{4}(4-2x_1-x_3)[/tex]

[tex]x_3=\frac{1}{5}(5-x_1-3x_2)[/tex]

First approximation,

[tex]x_1(1)=\frac{1}{8}(8-(0)-(0))=1[/tex]

[tex]x_2(1)=\frac{1}{4}(4-2(1)-(0))=0.5[/tex]

[tex]x_3(1)=\frac{1}{5}(5-1-3(0.5))=0.5[/tex]

Second approximation,

[tex]x_1(2)=\frac{1}{8}(8-(0.5)-(0.5))=0.875[/tex]

[tex]x_2(2)=\frac{1}{4}(4-2(0.875)-(0.5))=0.4375[/tex]

[tex]x_3(2)=\frac{1}{5}((0.875)-3(0.4375))=0.5625[/tex]

Third approximation,

[tex]x_1(3)=\frac{1}{8}(8-(0.4375)-(0.5625))=0.875[/tex]

[tex]x_2(3)=\frac{1}{4}(4-2(0.875)-(0.5625))=0.421875[/tex]

[tex]x_3(3)=\frac{1}{5}(5-(0.875)-3(0.421875))=0.571875[/tex]

Fourth approximation,

[tex]x_1(4)=\frac{1}{8}(8-(0.421875)-(0.571875))=0.875781[/tex]

[tex]x_2(4)=\frac{1}{4}(4-2(0.875781)-(0.571875))=0.419141[/tex]

[tex]x_3(4)=\frac{1}{5}(5-(0.875781)-3(0.419141))=0.573359[/tex]

Fifth approximation,

[tex]x_1(5)=\frac{1}{8}(8-(0.419141)-(0.573359))=0.875938[/tex]

[tex]x_2(5)=\frac{1}{4}(4-2(0.875938)-(0.573359))=0.418691[/tex]

[tex]x_3(5)=\frac{1}{5}(5-(0.875938)-3(0.418691))=0.573598[/tex]

Hence, by the Gauss Seidel method the solution of the given system is,

[tex]x_1\approx 0.876[/tex]

[tex]x_2\approx 0.419[/tex]

[tex]x_3\approx 0.574[/tex]

Answer:

4

Step-by-step explanation:

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: 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:

  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:

x = 25

Step-by-step explanation:

Step 1: Identify the similar triangles

Triangle DQC and triangle DBR are similar

Step 2: Identify the parallel lines

QC is parallel to BR

Step 3: Find x

DQ/QB = DC/CR

40/24 = x/15

x = 25

!!

Answer: [tex]x=25[/tex]

Step-by-step explanation:

In order to calculate the value of "x", you can set up de following proportion:

[tex]\frac{BQ+QD}{QD}=\frac{RC+CD}{CD}\\\\\frac{24+40}{40}=\frac{15+x}{x}[/tex]

Now, the final step is to solve for "x" to find its value.

Therefore, its value is the following:

[tex]1.6=\frac{15+x}{x}\\\\1.6x=x+15\\\\1.6x-x=15\\\\0.6x=15\\\\x=\frac{15}{0.6}\\\\x=25[/tex]

Answer:  Probability that she attended class regularly given that she receives A grade is 0.9286.

Step-by-step explanation:

Since we have given that

Probability of those who attend regularly receive A's in the class = 35%

Probability of those who do not regularly receive A's in the class = 5%

Probability of students who attend class regularly = 65%

We need to find the probability that she attended class regularly given that she receives an A grade.

Let E be the event of students who attend regularly.

P(E) = 0.65

And P(E') = 1-0.65 = 0.35

Let A be the event who attend receive A in the class.

So, P(A|E) = 0.35

P(A|E') = 0.05

So, According to question, we have given that

[tex]P(E|A)=\dfrac{P(E)P(A|E)}{P(E)P(A|E)+P(E')P(A|E')}\\\\P(E|A)=\dfrac{0.65\times 0.35}{0.65\times 0.35+0.35\times 0.05}\\\\P(E|A)=\dfrac{0.2275}{0.2275+0.0175}=\dfrac{0.2275}{0.245}=0.9286[/tex]

Hence, Probability that she attended class regularly given that she receives A grade is 0.9286.

Final answer:

The probability that a student attended class regularly given they received an A is approximately 0.9286, or 92.86% when rounded to four decimal places, calculated using Bayes' theorem.

Explanation:

To solve the problem, we need to calculate the conditional probability that a student attended class regularly given they received an A grade. To do this, we'll use Bayes' theorem, which allows us to reverse conditional probabilities.

Let's denote Attendance as the event that a student attends class regularly and A as the event of a student receiving an A grade. According to the question:

The overall probability of receiving an A, P(A), is computed as follows:

P(A) = P(A|Attendance) × P(Attendance) + P(A|Not Attendance) × P(Not Attendance)
    = 0.35 × 0.65 + 0.05 × (1 - 0.65)
    = 0.2275 + 0.0175
    = 0.2450

Now we use Bayes' theorem to find P(Attendance|A), the probability of attendance given an A:

P(Attendance|A) = (P(A|Attendance) × P(Attendance)) / P(A)
       = (0.35 × 0.65) / 0.245
       = 0.2275 / 0.245
       ≈ 0.9286

Therefore, the probability that a student attended class regularly given that they received an A grade is approximately 0.9286, or 92.86% when rounded to four decimal places.

Answer: Option(e) exist 123.50 is correct.

Step-by-step explanation:

James earns:

Product A: 15 × 35.75 = 536.25

Product B: 16 × 42.25 = 676

Total Earnings = 1212.25

Sally earns:

Product A: 25 × 35.75 = 893.75

Product B: 10 × 42.25 = 422.5

Total Earnings = 1316.25

Andre earns:

Product A: 18 × 35.75 = 643.5

Product B: 13 × 42.25 = 549.25

Total Earnings = 1192.75

Above calculation shows that Sally is the most profitable seller and Andre is the least profitable seller.

So, the difference between the most profitable seller i.e Sally (1316.25) and the least profitable seller i.e. Andre (1192.75) is 123.50.

Answer: 1.64

Step-by-step explanation:

80% = 50

20% = 12.5

100% = 62.5

38% = 1.64

Answer:x=-3,-2,3

Step-by-step explanation:

Given equation of polynomial is

[tex]x^{3}+2x^2-9x-18=0[/tex]

taking [tex]x^3[/tex] and -9x together and remaining together we get

[tex]x^3-9x+2x^2-18=0[/tex]

[tex]x\left ( x^2-9\right )+2\left ( x^2-9\right )[/tex]

[tex]x\left ( \left ( x+3\right )\left ( x-3\right )\right )+2\left ( \left ( x+3\right )\left ( x-3\right )\right )[/tex]

[tex]taking \left ( x+3\right )\left ( x-3\right ) as common[/tex]

[tex]\left ( x+2\right )\left ( x+3\right )\left ( x-3\right )=0[/tex]

therefore

x=-3,-2,3

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.

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:

Step-by-step explanation:

.5

Answer is provided in image attached.

Answer:

  1500 ft²

Step-by-step explanation:

The sum of two adjacent sides of the pasture is half the perimeter (160 ft/2 = 80 ft), so the side adjacent to the 50 ft side will be 80 ft - 50 ft = 30 ft.

The product of adjacent sides of a rectangle gives the area of the rectangle. That area will be ...

  area = (50 ft)(30 ft) = 1500 ft²

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:  [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:

75.8%

Step-by-step explanation:

Mean weight of chickens = u = 1700 g

Standard deviation = [tex]\sigma[/tex] = 200g

We need to calculate the percentage of chickens having weight more than 1560 g

So,

x = 1560 g

Since the weights can be approximated by normal distribution, we can use concept of z-score to solve this problem.

First we need to convert the given weight to z score. The formula for z score is:

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

Using the values, we get:

[tex]z=\frac{1560-1700}{200} \\\\ z = -0.7[/tex]

So now we have to calculate what percentage of values lie above the z score of -0.7. Using the z-table or z-calculator we get:

P(z > -0.7) = 0.758

This means 0.758 or 75.8% of the values are above z score of -0.7. In context of our question we can write:

75.8% of the chickens will have weight more than 1560 g

To find the percent of chickens having weights more than 1560 g, calculate the z-score for 1560 g and find the area to the right of this z-score in the standard normal distribution curve.

To find the percent of all chickens having weights more than 1560 g, we need to calculate the z-score for 1560 g and then find the area to the right of this z-score in the standard normal distribution curve.

First, calculate the z-score using the formula: z = (x - μ) / σ, where x is the weight of the chicken, μ is the mean weight, and σ is the standard deviation.

For the weight 1560 g, the z-score is calculated as: z = (1560 - 1700) / 200 = -0.7

Using a standard normal distribution table or calculator, find the area to the right of -0.7. This area represents the percent of chickens having weights more than 1560 g.

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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:

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: 0.2401

Step-by-step explanation:

The binomial distribution formula is given by :-

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

where P(x) is the probability of x successes out of n trials, p is the probability of success on a particular trial.

Given : The probability that a randomly chosen citizen-entity of Cygnus is of pension age† is approximately: p =0.7.

Number of trials  : n= 4

Now, the required probability will be :

[tex]P(x=4)=^4C_4(0.7)^4(1-0.7)^{4-4}\\\\=(1)(0.7)^4(1)=0.2401[/tex]

Thus, the probability that, in a randomly selected sample of four citizen-entities, all of them are of pension age =0.2401

The norm of a vector [tex]\vec x[/tex] is equal to the square root of the inner product of [tex]\vec x[/tex] with itself.

a. [tex]\|(-1,2)\|=\sqrt{\langle(-1,2),(-1,2)\rangle}=\sqrt{4(-1)^2+9(2)^2}=\sqrt{40}=2\sqrt{10}[/tex]

b. [tex]\|(3,2)\|=\sqrt{\langle(3,2),(3,2)\rangle}=\sqrt{4(3)^2+9(2)^2}=\sqrt{72}=6\sqrt2[/tex]

Answer:

• (x, y, z) = (3+t, 1-t, 8+4t) . . . equation of the line

• xy-intercept (1, 3, 0)

• yz-intercept (0, 4, -4)

• xz-intercept (4, 0, 12)

Step-by-step explanation:

The line's direction vector is given by the coordinates of the plane: (1, -1, 4). So, the parametric equations can be ...

(x, y, z) = (3, 1, 8) + t(1, -1, 4) . . . . . parametric equation for the line

or

(x, y, z) = (3+t, 1-t, 8+4t)

__

The various intercepts can be found by setting the respective variables to zero:

xy-plane: z=0, so t=-2. (x, y, z) = (1, 3, 0)

yz-plane: x=0, so t=-3. (x, y, z) = (0, 4, -4)

xz-plane: y=0, so t=1. (x, y, z) = (4, 0, 12)

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: 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.)

I suppose

[tex]H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}[/tex]

The vectors that span [tex]H[/tex] form a basis for [tex]P_2[/tex] if they are (1) linearly independent and (2) any vector in [tex]P_2[/tex] can be expressed as a linear combination of those vectors (i.e. they span [tex]P_2[/tex]).

Compute the Wronskian determinant:

[tex]\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0[/tex]

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of [tex]H[/tex] is 3.

Write an arbitrary vector in [tex]P_2[/tex] as [tex]ax^2+bx+c[/tex]. Then the given vectors span [tex]P_2[/tex] if there is always a choice of scalars [tex]k_1,k_2,k_3[/tex] such that

[tex]k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c[/tex]

which is equivalent to the system

[tex]\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}[/tex]

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

[tex]\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}[/tex]

so the vectors do span [tex]P_2[/tex].

The vectors comprising [tex]H[/tex] form a basis for it because they are linearly independent.

To determine if a set of polynomials forms a basis for P2, they need to be linearly independent and span the vector space P2. If the only solution to a homogeneous system of equations is trivial (all coefficients equal zero), they are linearly independent. Whether they span P2 or not depends on if any polynomial of degree 2 or less can be expressed as a linear combination of these polynomials.

In order to determine if the set of polynomials {10x2+4x, 3x-4x2+3, 5x2+x} forms a basis for P2, we need to prove two properties: they should be linearly independent and they should span the vector space P2.

Linear independence means that none of the polynomials in the given set can be expressed as a linear combination of the others. The simplest way to prove this is to set up a system of equations called a homogeneous system, and solve for the coefficients. If the only solution to this system is the trivial solution (where all coefficients equal zero), then they are linearly independent.

Spanning means that any polynomial of degree 2 or less can be expressed as a linear combination of these polynomials.

So, depending on the outcome of checking those two properties, we can determine if the given set of polynomials is a basis for P2 or not.

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