What fraction in this list is more than 3/5? 20/100, 6/10, 1/2, 2/12 or 2/3?

To find the limit of the given function as t approaches infinity, we'll analyze the function:

lim (t → ∞) (t + t^2) / (5t - t^2)

Step 1: Factor out the highest power of t in the numerator and the denominator.

t^2(1/t + 1) / t^2(-1 + 5/t)

Step 2: Simplify the expression by canceling out the t^2 terms.

(1/t + 1) / (-1 + 5/t)

Step 3: Evaluate the limit as t approaches infinity.

lim (t → ∞) (1/t + 1) / (-1 + 5/t)

As t approaches infinity, the terms 1/t and 5/t both approach 0.

(0 + 1) / (-1 + 0)

Step 4: Simplify the expression to find the limit.

1 / -1 = -1

So, the limit of the given function as t approaches infinity is -1.

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The force that is applied to move the object is obtained as 25.4 N

Let us recall that in physics work is the product of the force and the distance that is covered.

W = F * d

where W is the amount of work completed, F is the force applied, d is the distance traveled, and theta is the angle formed by the force's and the displacement's directions.

Thus we have;

Work = Force * distance

Force = Work/Distance

= 350J/13.8 m

= 25.4 N

The force that is applied is 25.4 N

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The solution to each of the compound interest problems are:

A = 9000(1.0425)^(t)

B) t = 19.2 years

C)  $72118.34

The formula for the compound interest here is:

A = P(1 + (r/n))^nt

where:

A = Accrued amount (principal + interest)

P = Principal amount

r = Annual nominal interest rate as a decimal

R = Annual nominal interest rate as a percent

r = R/100

n = number of compounding periods per unit of time

t = time in decimal years; e.g., 6 months is calculated as 0.5 years. Divide your partial year number of months by 12 to get the decimal years.

I = Interest amount

We are given:

P = $9000

r = 4.25% = 0.0425

n = 1

a) A = 9000(1 + 0.0425)^(t)

A = 9000(1.0425)^(t)

B) For the account to exceed $20000, we have:

20000 = 9000(1.0425)^(t)

20/9 = (1.0425)^(t)

t In 1.0425 = In (20/9)

t = 19.2 years

C) For t = 50 years, we have:

9000(1.0425)^(50) = $72118.34

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The value of x is 4 from the given figure with circle

We have to find the value of x in the given figure

3(3+5)=x(2+x)

Apply distributive property

24=2x+x²

x² +2x-24=0

x² +6x-4x-24=0

x(x+6)-4(x+6)=0

(x-4)(x+6)=0

x=4 or x=-6

The value should be positive

Hence, the value of x is 4 from the given figure with circle

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The location where Jim should park to minimize the distance he must walk from his car to the donut shop is approximately (10/9, 16/3).

To find the location where Jim should park to minimize the distance he must walk from his car to the donut shop, we can use the concept of the perpendicular bisector.

Step 1: Find the midpoint of the line segment connecting the two points (0,1) and (4,3). The midpoint can be found by taking the average of the x-coordinates and the average of the y-coordinates, i.e.,

Midpoint = ( (0+4)/2 , (1+3)/2 ) = (2,2)

Step 2: Find the slope of the line connecting the two points (0,1) and (4,3). The slope can be found using the formula

slope = (y2 - y1) / (x2 - x1)

where (x1,y1) = (0,1) and (x2,y2) = (4,3). Therefore,

slope = (3-1)/(4-0) = 1/2

Step 3: Find the equation of the perpendicular bisector of the line segment connecting the two points (0,1) and (4,3). The perpendicular bisector has a slope that is the negative reciprocal of the slope of the line segment, which is -2. The equation of the perpendicular bisector passing through the midpoint (2,2) can be found using the point-slope form of a linear equation,

y - y1 = m(x - x1)

where m is the slope and (x1,y1) is the midpoint. Therefore, the equation of the perpendicular bisector is

y - 2 = -2(x - 2)

Simplifying this equation gives

y = -2x + 6

Step 4: Find the point on the line y = -2x + 6 that is closest to the point (2,4), which is the location of the donut shop. The distance between the point (2,4) and any point on the line y = -2x + 6 can be found using the distance formula,

distance = sqrt( (x - 2)^2 + (y - 4)^2 )

To minimize this distance, we can minimize the squared distance,

distance^2 = (x - 2)^2 + (y - 4)^2

Using the equation of the line y = -2x + 6, we can substitute y = -2x + 6 into the equation for the squared distance to get

distance^2 = (x - 2)^2 + (-2x + 2)^2

Taking the derivative of distance^2 with respect to x and setting it equal to zero gives the critical point,

d(distance^2)/dx = 2(x - 2) + 2(-2x + 2)(-2) = 0

Solving for x gives

x = 10/9

Substituting x = 10/9 into the equation for the line y = -2x + 6 gives

y = -2(10/9) + 6 = 16/3

Therefore, the location where Jim should park to minimize the distance he must walk from his car to the donut shop is approximately (10/9, 16/3).

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The minimum area of board he should use to distribute his weight is 1.6 m^2

We have to note that the pressure is the ratio of the force and the area of the object. In this case, we can see that we have the force of the object in the form of its weightg which is given in the question as g on. The roofer has a weight of  while the pressure has been given in the question that we have as  500 N/m².

Knowing that;

Pressure = Force/Area

Area = Force/Pressure

Area = 880 N/550N/m^2

Area = 1.6 m^2

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The Penders' total expenditure for the month of July is $126.45 less than the amount they had budgetd.

To solve for total expenditure, we total the amount of money spent for the month and minus it by the total amount of money budgeted for the month. Or you some the balance.

9. How much more or less were the Penders' total expenditures for the month of July than the amount they had budgeted? (2 points)

Expense Summary for Paul and Diana Pender for the Month of July 20                

                                        Amount B.     Amount S.        Difference

(Food/Groceries)

Food/Groceries               $290.00          $302.60       a. -12.6

(Household Expenses)    

Electricity                          $45.00            $44.35         b. 0.65

Heating                             $80.00             $0.00          c   80

Cell Phone                      $35.00              $35.00         d.   0

Water                               $24.50              $31.70          e.   -7.20

Cable/Internet                 $95.00              $95.00         f       0

(Transportation)

Gasoline/Oil                    $85.00               $101.70      g     -16.7

Parking/Tolls                    $70.00             $50.00        h.     20

(Personal)

Clothing                           $60.00                $31.75          i   28.25

Credit Card(s)                 $50,00                 $60.00          j   -10

Pocket Money                $80.00                 $93.75          k.   -13.75

(Entertainment)

Movies/Theater              $20.00                $35.00          L   -15

Sporting Events              $65.00                 $32.00         m.  33

Recreation                     $22.00                   $63.80         n.  -41.80

Dining Out                     $140.00                 $158.40        o.    -18.40

(Fixed)

Rent/Mortgage              $625.00               $625.00        p.  0

Furniture                      $125.00                  $125.00         q 0

Savings                          $250.00                $150.00        r  100

Contributions                 $8.33                       $8.33          S.  0                

                                      $2169.83               $2042.38

$290 + $45 + $80 + $35 + $24.50 + $95 + $85 + $70 + $60 + $50 + $80 + $20 + $65 + $22 + $140 + $625 + $125 + $250 + $8.33 = 2169.83

$302.60 + $44.35 + $0 + $35 + $31.70 + $95 + $101.70 + $50 + $31.75 + $60 + $93.75 + $35 + $32 + $63.80 + $158.40 + $625 + $125 + $150 + $8.33 = 2043. 38

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Answer:

210

Step-by-step explanation:

you just need to multiply all numbers

a) The probability that only two of Taylor's photos will be chosen as finalists is (10 * 1,218,336) / 4,496,388, which is approximately 0.027., b) The probability that at least 4 of Taylor's photos will be chosen as finalists is 0.07.

A) To calculate the probability that only two of Taylor's photos will be chosen as finalists, we need to use the combination formula. The total number of ways to choose 8 photos out of 40 is 40 choose 8, which is approximately 4,496,388. The number of ways to choose exactly 2 of Taylor's photos is 5 choose 2, which is 10. The number of ways to choose the remaining 6 photos out of the 35 that are not Taylor's is 35 choose 6, which is approximately 1,218,336. So, the probability that only two of Taylor's photos will be chosen as finalists is (10 * 1,218,336) / 4,496,388, which is approximately 0.027.

B) To calculate the probability that at least 4 of Taylor's photos will be chosen as finalists, we need to consider all the possible ways that this could happen. There are several cases to consider:
- 4 of Taylor's photos are chosen and 4 photos from the other submissions are chosen
- 5 of Taylor's photos are chosen and 3 photos from the other submissions are chosen
- 6 of Taylor's photos are chosen and 2 photos from the other submissions are chosen
- 7 of Taylor's photos are chosen and 1 photo from the other submissions is chosen
- All 8 finalists are Taylor's photos

We can calculate the probability of each case separately and then add them up to get the total probability.

For the first case, the number of ways to choose 4 of Taylor's photos is 5 choose 4, which is 5. The number of ways to choose 4 photos from the other submissions is 35 choose 4, which is approximately 52,360. The probability of this case is (5 * 52,360) / 4,496,388, which is approximately 0.058.

For the second case, the number of ways to choose 5 of Taylor's photos is 5 choose 5, which is 1. The number of ways to choose 3 photos from the other submissions is 35 choose 3, which is approximately 52,360. The probability of this case is (1 * 52,360) / 4,496,388, which is approximately 0.012.

For the third case, the number of ways to choose 6 of Taylor's photos is 5 choose 6, which is 0. There are no ways to choose 2 photos from the other submissions since there are only 3 left. So the probability of this case is 0.

For the fourth case, the number of ways to choose 7 of Taylor's photos is 5 choose 7, which is 0. There are no ways to choose 1 photo from the other submissions since there are only 2 left. So the probability of this case is 0.

For the fifth case, the probability is simply (5 choose 8) / (40 choose 8), which is approximately 0.000003.

Adding up the probabilities from all the cases, we get a total probability of approximately 0.07. So the probability that at least 4 of Taylor's photos will be chosen as finalists is 0.07.
A) The probability that only two of Taylor's photos will be chosen as finalists from the 40 submissions is calculated using the formula: (number of ways to choose 2 photos from Taylor's 5 submissions) * (number of ways to choose 6 photos from the remaining 35 submissions) / (number of ways to choose 8 photos from the total 40 submissions). This can be calculated as (5C2 * 35C6) / 40C8.

B) The probability that at least 4 of Taylor's photos will be chosen as a finalist can be calculated by finding the probability of 4, 5, or all of Taylor's photos being chosen and then adding these probabilities together. The individual probabilities can be calculated in a similar manner to part A. The final answer will be the sum of these individual probabilities.

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D. 12 persons were among the walkers who got more than 30 pledges.

From the data table there are a total of 13 walkers, when the frequencies are summed up: (1 + 7 + 4 + 1 = 13).

Pledges that were received by:

40 and 49 = 7 walkers (a frequency of 7 for that interval),

50 and 59 = 4 walkers  (a frequency of 4 for that interval), and

60 and 69 = 1 walker (a frequency of 1 for that interval).

So the total number of walkers who received more than 39 pledges is 7 + 4 + 1 = 12.

Therefore, the number of walkers that received more than 30 pledges is 12 people.

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Complete question:

Kianna and her friends collected pledges for a local walk-a-thon.

They recorded the number of pledges they each received in a frequency table.

Chart with Pledges and their Frequency. 30 to 39 has frequency as 1, 40 to 49 has frequency as 7, 50 to 59 has frequency as 4 and 60 to 69 has frequency as 1.

How many walkers received more than 39 pledges?

A. 13

B. 11

C. 7

D. 12

To show that if one set is a subset of another then the other's complement is a subset of the first's complement, we can use the definition of set complement and subset.

Let's assume that A is a subset of B, which means that every element of A is also an element of B. We want to show that the complement of B, denoted by B', is a subset of the complement of A, denoted by A'.

To prove this, we need to show that every element of B' is also an element of A'. Let x be an arbitrary element of B'. By definition, x is not an element of B. Since A is a subset of B, x cannot be an element of A either. Therefore, x must be an element of A', which means that B' is a subset of A'.

In summary, if A is a subset of B, then B' is a subset of A'.

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The number of feet that the player A must run to reach the home team's basket using the distance formula is 64.5 feet to the nearest tenth of a feet.

Given a coordinate grid which shows the movement of the player A on a basketball court using the coordinate grid.

Position of player A = (-3, 2)

Home team's basket is at (6, 0)

We have to find the distance between the two points.

using the distance formula,

Distance between the home team's basket and player A is,

Distance = √(6 - -3)² + (0 - 2)²

               = √(9² + 2²)

               = √85 units

Given that each unit is 7 feet.

Distance = 7 × √85 units = 64.5 feet

Hence the distance from player A to the basket is 64.5 feet.

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The circumference of the circle is 8π cm.

Given the diameter of the circle as 8cm,

The radius (r) can be gotten by dividing the diameter by 2

r = 8cm / 2 = 4cm

The area (A) of a circle is: A = πr^2

So, substituting the value of r, we get:

A = π(4cm)^2 = 16π cm^2

Therefore, the area of the circle is 16π cm^2.

The circumference (C) of a circle is C = 2πr

So, substituting the value of r, we get:

C = 2π(4cm) = 8π cm

Therefore, the circumference of the circle is 8π cm.

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a. The first quarter's interest for the deposit that Bailey makes in a savings account paying 4¹/₂% interest compounded quarterly is $81.00.

b. The first quarter's balance is $7,281.00.

c. The second quarter's interest for the deposit that Bailey makes in a savings account paying 4¹/₂% interest compounded quarterly is $81.91.

d. The second quarter's balance is $7,362.91.

e. The third quarter's interest for the deposit that Bailey makes in a savings account paying 4¹/₂% interest compounded quarterly is $82.83.

f. The third quarter's balance is $7,445.74.

g. The first quarter's balance is $7,529.51.

h. The fourth quarter's interest for the deposit that Bailey makes in a savings account paying 4¹/₂% interest compounded quarterly is $83.76.

i. The total interest earned in the first year by the account is $329.51.

The interest type is the compound interest system.

The compound interest system charges interest on accumulated interest and the principal, unlike the simple interest system that charges interest on only the principal for each period.

N (# of periods) = 4 quarters (1 year x 4)

I/Y (Interest per year) = 4¹/₂% = 4.5%

PV (Present Value) = $7,200

PMT (Periodic Payment) = $0

Results:

FV = $7,529.51

Total Interest = $329.51

Period     PV                  PMT          Interest                FV

1         $7,200.00         $0.00          $81.00         $-7,281.00

2         $7,281.00         $0.00          $81.91          $-7,362.91

3         $7,362.91         $0.00         $82.83         $-7,445.74

4         $7,445.74        $0.00          $83.76         $-7,529.51

Year #1 end

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The eigenbasis for problem 21 is {[0, 0, 1]}.

To find the eigenvalues, we solve the characteristic equation:

|A - λI| = 0

where A is the matrix and I is the identity matrix of the same size.

For problem 21, we have:

A = [1 0 0; -5 0 2; 0 0 1]

I = [1 0 0; 0 1 0; 0 0 1]

So,

|A - λI| = det([1-λ 0 0; -5 0 2; 0 0 1-λ])

= (1-λ) det([0 2; 0 1-λ]) + 5 det([-5 2; 0 1-λ])

= (1-λ)(1-λ)(-5) + 5(-10)

= 25λ - 125

= 25(λ - 5)

Thus, the only eigenvalue is λ = 5.

To find the eigenvectors, we solve the system of equations:

(A - λI)x = 0

For λ = 5, we have:

(A - λI)x = [(1-5) 0 0; -5 (0-5) 2; 0 0 (1-5)]x = [-4 0 0; -5 -5 2; 0 0 -4]x = 0

This gives us the system of equations:

-4x1 = 0

-5x1 - 5x2 + 2x3 = 0

-4x3 = 0

From the first and third equations, we see that x1 = 0 and x3 = 0. Then the second equation reduces to:

-5x2 = 0

So, we have x2 = 0. Thus, the eigenspace for λ = 5 is spanned by the vector [0, 0, 1].

Since we only have one eigenvalue, we automatically have an eigenbasis. So, the eigenbasis for problem 21 is {[0, 0, 1]}.

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The probability that two batteries used sequentially will last more than four years is approximately 0.0835.

We recognize that the time to failure of a rechargeable battery is exponentially distributed with a mean of 3 years. consequently, the parameter lambda for the exponential distribution is:

lambda = 1/mean = 1/3

Let X1 be the time to failure of the first battery and X2 be the time to failure of the second battery. We want to find the possibility that each batteries last more than four years, which may be expressed as:

P(X1 > 4 and X2 > 4)

The use of the memoryless belongings of the exponential distribution, we will rewrite this chance as:

P(X1 > 4) x P(X2 > 4)

The opportunity density feature of an exponential distribution with parameter lambda is:

[tex]f(x) = lambda * e^{(-lambda*x)}, for x > = 0[/tex]

Therefore, the opportunity that a battery lasts more than four years is:

P(X > 4) = imperative from 4 to infinity of lambda * [tex]e^{(-lambdax)}[/tex] dx

= [tex]e^{(-lambda4)}[/tex]

Substituting lambda = 1/3, we get:

P(X > 4) = [tex]e^{(-4/3)}[/tex]

The use of the memoryless property, we've got:

P(X1 > 4) =[tex]e^{(-4/3)}[/tex]

P(X2 > 4) = [tex]e^{(-4/3)}[/tex]

Consequently, the chance that both batteries last more than 4 years is:

P(X1 > 4 and X2 > 4) = P(X1 > 4) x P(X2 > 4)

[tex]= e^{(-4/3)} x e^{(-4/3)}[/tex]

[tex]= e^{(-8/3)}[/tex]

≈ 0.0835

Consequently, the probability that two batteries used sequentially will last more than four years is approximately 0.0835.

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7-14 are similar methods to find the general solutions of the systems whose augmented matrices are given in exercises

To find the general solutions of the systems whose augmented matrices are given in exercises 7-14, we need to perform row operations until the augmented matrix is in row echelon form or reduced row echelon form. Then, we can use back substitution to solve for the variables.

The term "solutions" refers to the set of values that satisfy the system of equations represented by the augmented matrix. The term "augmented" refers to the matrix formed by appending the column vector of constants to the coefficient matrix.

Once we have the augmented matrix in row echelon form or reduced row echelon form, we can identify the pivot variables and free variables. Pivot variables are the variables corresponding to the pivot columns, while free variables are the remaining variables. We can express the pivot variables in terms of the free variables to obtain the general solution.

For example, consider the following augmented matrix:

[1 2 -1 | 0]
[2 4 1 | 5]
[-1 1 2 | -1]

To put this matrix in row echelon form, we can perform the following row operations:

R2 - 2R1 -> R2
R3 + R1 -> R3
R3 + 2R2 -> R3

This gives us the following row echelon form:

[1 2 -1 | 0]
[0 0 3 | 5]
[0 0 0 | 4]

The pivot variables are x1 and x3, while x2 is a free variable. We can express x1 and x3 in terms of x2 as follows:

x1 = -2x2
x3 = 5/3 - (5/3)x2

Therefore, the general solution is:

x1 = -2x2
x2 = x2
x3 = 5/3 - (5/3)x2

This can be written more compactly as:

x = [-2x2, x2, 5/3 - (5/3)x2]

where x is the vector of variables.

We can apply similar methods to find the general solutions of the systems whose augmented matrices are given in exercises 7-14.

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Find the general solutions of the systems whose augmented ma- trices are given in Exercises 7-14. 10. B -2 -1 3-6-2 2]

The amount of brown sugar called for in the recipe is 1134.0 grams, tenths of a gram rounded up. By dividing 40 ounces by the conversion value of 28.35 grams per ounce, you can calculate this and get 1134.0 grams.

Rounding to the closest tenth of a gram is necessary to convert 40 ounces to grams. To do this, multiply the weight in ounces by the conversion factor of 28.35 (1 ounce = 28.35 grams).

Therefore, 40 ounces of brown sugar are equivalent to:

40 ounces x 28.35 grams per ounce = 1134 grams

Because 1 gram only roughly equates to 0.04 ounces, it is vital to keep in mind that this is an estimation. This conversion factor may change based on the situation and the material being measured.

Hence, The amount of brown sugar required for the recipe is roughly 1134 grams.

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There are 8,222,286 different ways that six students can form a group for an activity in a class of 29 students.

To find the number of ways to form a group of 6 students from a class of 29, we can use the concept of combinations, which are arrangements of items without regard to order.

The number of combinations of n items taken r at a time is denoted by C(n,r) and is given by the formula:

C(n,r) = n! / (r!(n-r)!)

where n! denotes the factorial of n, which is the product of all positive integers from 1 to n. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

To apply this formula to the problem, we have:

n = 29 (number of students in the class)

r = 6 (number of students in the group)

So, the number of ways to form a group of 6 students from a class of 29 is:

C(29,6) = 29! / (6! * (29-6)!) = (29 x 28 x 27 x 26 x 25 x 24) / (6 x 5 x 4 x 3 x 2 x 1) = 8,222,286

Therefore, there are 8,222,286 different ways that six students can form a group for an  activity in a class of 29 students.

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The measures of the indicated angles will be 80°, 44°, and 93°.

It should be noted that a triangle is a three-sided polygon that consists of three edges and three vertices. The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees.

For triangle TUV, the value will be:

= 180 - 63 - 37

= 80°

For triangle ABC, the value will be:

= 180 - 90 - 46

= 44.

For triangle PQR the value will be:

= 180 - 51 - 36

= 93

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The disc jockey could use a Monte Carlo simulation to estimate the probability that the next three songs are all rock songs.

In this simulation, she would randomly select a rock or pop song for each of the three slots, and then repeat this process many times (e.g. 10,000 times). She could then count the number of times that all three songs were rock songs, and divide this by the total number of simulations to get an estimate of the probability.

To estimate the probability that the next three songs are all rock songs, the disc jockey could use the following simulation design:
1. Assign a number to each rock and pop song, ensuring that both genres have equal numbers.
2. Use a random number generator to select three numbers corresponding to the songs.
3. Record the genres of the chosen songs and note if all three are rock songs.
4. Repeat the simulation process multiple times (e.g., 1000 times) to obtain a larger sample.
5. Calculate the probability by dividing the number of times all three selected songs were rock songs by the total number of simulations performed.
This simulation design will help estimate the probability of the next three songs being rock songs by accounting for the equal number of rock and pop songs in the selection pool.

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Answer:

2³×5²

Step-by-step explanation:

200 as a product of prime factors is 2×2×2×5×5, which is 2³×5²

Answer: There are different ways to approach this problem, but one possible method is to use the concept of percentile rank and cumulative distribution function (CDF) of the data.

Assuming we have a data set of distances that residents on this street traveled to work in a week, we can first calculate the CDF of the data, which gives the probability of observing a value less than or equal to a certain distance. For example, if there are 100 residents and 20 of them traveled 50 miles or less, then the CDF at 50 miles is 20/100 = 0.2.

Once we have the CDF, we can find the percentile rank of a given distance by multiplying the CDF by 100. For example, if the CDF at 85 miles is 0.75, then the percentile rank of a resident who traveled 85 miles is 0.75 x 100 = 75%.

Since we do not have the actual data, we cannot calculate the CDF directly. However, we can make some assumptions and estimates based on the information given. For example, if we assume that the distribution of distances is roughly normal with a mean of 30 miles and a standard deviation of 15 miles, then we can use the properties of the standard normal distribution to estimate the percentile rank of 85 miles.

Specifically, we can standardize the distance by subtracting the mean and dividing by the standard deviation:

z = (85 - 30) / 15 = 3.33

Then, we can use a standard normal distribution table or calculator to find the percentile rank of z, which is the same as the percentile rank of 85 miles in this distribution.

According to the table or calculator, the area under the standard normal curve to the left of z = 3.33 is approximately 0.9993. This means that about 99.93% of the distances in this distribution are less than 85 miles. Therefore, the percentile rank of a resident who traveled 85 miles is approximately 100 - 99.93 = 0.07, or 7%.

Since none of the answer choices match this result exactly, we can choose the closest option, which is (b) 70.

To find the largest number of vertices for a simple undirected graph with 17 edges and each vertex of at least degree 3, we can use the handshaking lemma: the sum of degrees of all vertices in a graph is twice the number of edges. Therefore, if there are V vertices in the graph, the sum of degrees of all vertices is at least 3V. So we have:

2 * 17 = sum of degrees of all vertices ≥ 3V

34 ≥ 3V

V ≤ 11.33

Therefore, the largest number of vertices this graph can have is 11 (since it must be a whole number).

To give an example of such a graph with maximal number of vertices, we can construct a graph with 11 vertices, each with degree 3, and 17 edges connecting them. One possible graph that satisfies these conditions is a regular icosahedron, which has 12 vertices and 30 edges, but we can remove one vertex and three edges to obtain the desired graph.

To prove that there is not such graph with a larger amount of vertices, we can use a contradiction argument. Suppose there is a simple undirected graph with 18 or more vertices, each of at least degree 3, and 17 edges. By the handshaking lemma, the sum of degrees of all vertices in this graph is at least 3 times the number of vertices, which is at least 54. However, there are only 17 edges in the graph, which means that the sum of degrees of all vertices is at most 2 times the number of edges, which is 34. This is a contradiction, since 34 is less than 54. Therefore, there is no such graph with a larger amount of vertices.

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The correct answer is option D) 22,25,33,35,37,40,45.

To create a box plot, we need to find the minimum value, the maximum value, the median, and the quartiles of the data set.

Now, let's check which of the given data sets matches the information provided in the box plot:

A) 22,29,34,36,39,45: This data set does not have a minimum value of 20 or a maximum value of 45, so it cannot be represented by the given box plot.

B) 20,30,35,40,45: This data set has the correct minimum and maximum values, but it does not have a median of 32.5 or an IQR of 10, so it cannot be represented by the given box plot.

C) 22,25,35,35,35,35,36,44,45: This data set has the correct minimum and maximum values and a median of 32.5, but it does not have an IQR of 10. The Q₁ and Q₃ values are also different from the given box plot, so it cannot be represented by the given box plot.

D) 22,25,33,35,37,40,45: This data set has the correct minimum and maximum values, a median of 32.5, and an IQR of 10. The Q1 and Q3 values also match the given box plot. Therefore, this data set could be represented by the given box plot.

So, the correct answer is option D) 22,25,33,35,37,40,45.

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For Felipe and Helena's final grades, the solution is option C, [74 71].

Using the given values for Q, T, and P weights and Felipe and Helena's grades, calculate their final grades as follows:

Felipe's final grade:

0.40 x 80 + 0.50 x 60 + 0.10 x 90 = 32 + 30 + 9 = 71

Helena's final grade:

0.40 x 70 + 0.50 x 80 + 0.10 x 60 = 28 + 40 + 6 = 74

To represent the final grades for Felipe and Helena in a matrix F, given formula F = WG, where W = matrix of weights and G = matrix of grades:

[0.40 0.50 0.10]   [80 70]

F = WG = [0.40 0.50 0.10] x [60 80]

[0.40 0.50 0.10] [90 60]

Performing matrix multiplication:

[32 + 30 + 9  28 + 40 + 6]

F = WG = [32 + 40 + 6 28 + 40 + 3]

[36 + 25 + 6 36 + 20 + 3]

Simplifying:

[71 74]

F = WG = [78 71]

[67 59]

Therefore, [74 71] for Felipe and Helena's final grades, respectively.

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thanks, this is a answer (:

An amount of $17.63 was worth 125.93 yuan in 2017 when 1 yuan was worth 0.14 U.S dollar.

We must divide the dollar amount by the exchange rate of yuan to dollars on that date to get the worth of yuan in 17.63 dollars on March 8, 2017

Convert 17.63 dollars to yuan. We get:

= 17.63 * (1/ 0.14)

= 125.928571429

= 125.93 yuan

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Estimating the quotient 6,581 divided by 77 gives 66

From the question, we have the following parameters that can be used in our computation:

6,581 divided by 77 estimation

To estimate the number is to approximate the number

Estimating 6581, we have 6600

Estimating 77, we have 100

This means that

6,581 divided by 77 estimation = 6600/100

Evaluate the quotient

6,581 divided by 77 estimation = 66

Hence, the estimate is 66

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