Ask a question about your assignment

The correct answer is C. Airport S has 100 on-time flights, and Airport T has 200 on-time flights. Since 100 is one-half of 200, Airport S has one-half the number of on-time flights that Airport T has.

Given that,

The number of flights that were either on time or delayed at three separate airports on a given day is depicted in the segmented bar chart below.

Which of the following statements is the bar chart evidence for?

The options are : In comparison to the other two airports, A Airport T has the highest proportion of on-time flights.

In comparison to the other two airports, B Airport R has the lowest percentage of on-time flights.

C The proportion of on-time flights at Airport S is less than half that at Airport T.

D The proportion of on-time flights at Airport R is lower than that at Airport S.

E There are exactly as many flights at Airport T as there are at Airports R and S put together.

As the number of flights that were either on time or delayed at three different airports on one day can be exactly represented by

The correct answer is less than half as many flights arrive on time at Airport S than at Airport T.

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The Complete Question is :

Which of the following statements is the bar chart evidence for the number of flights that were either on time or delayed at three different airports on one day.?

The given function f is increasing over the interval 0 < x < 2 and

decreasing over the interval 2 < x < 5.

What is a function?

A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets; mapping from A to B will be a function only when every element in set A has one end, only one image in set B.

An example of a simple function is f(x) = x^2. In this function, the function f(x) takes the value of “x” and then squares it. For instance, if x = 3, then f(3) = 9. A few more examples of functions are: f(x) = sin x, f(x) = x^2 + 3, f(x) = 1/x, f(x) = 2x + 3, etc.

Now,

As the given function f is increasing till x<2 and increasing after x>2

Hence,

            f is increasing over the interval 0 < x < 2 and

decreasing over the interval 2 < x < 5.

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Answer:

9.44

Step-by-step explanation:

7.19+2.25 =9.44

A dynamic array is an array with a big improvement: automatic resizing.

Explanation :

Initial Values:

n = 2

arr[0] = [ ]

arr[1] = [ ]

Query 0: Append 5 to arr[( 0 ⊕ 0 ) % 2 )] = arr[0]

last answer = 0

arr[0] = [5]

arr[1] = [ ]

Query 1: Append 7 to arr[( 1 ⊕ 0 ) % 2 )] = arr[1]

arr[0] = [5]

arr[1] = [7 ]

Query 2: Append 3 to arr[( 0 ⊕ 0 ) % 2 )] = arr[0]

last answer = 0

arr[0] = [5, 3]

arr[1] = [7 ]

Query 3: Assign the value at index 0 to arr [( 1 ⊕ 0 ) % 2 )] = arr[1] to

Last Answer, print Last Answer.

last answer = 7

arr[0] = [5, 3]

arr[1] = [7 ]

Query 4: Assign the value at index 1 to arr [( 1 ⊕ 7 ) % 2 )] = arr[0] to

Last Answer, print Last Answer.

last answer = 3

arr[0] = [5, 3]

arr[1] = [7 ]

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An identity is an equation that is always true for any value of the variable. To verify that an equation is an identity, we can substitute different values for the variable and see if the equation still holds true.

The given equation is (5sinx+5cosx)^2=25sin2x+25

We can start by expanding the left side of the equation:

(5sinx+5cosx)^2 = (5sinx)^2 + 2(5sinx)(5cosx) + (5cosx)^2

= 25sin^2x + 50sinxcosx + 25cos^2x

= 25(sin^2x + cos^2x) + 50sinxcosx

= 25 + 50sinxcosx

Now we can use the trigonometric identity sin^2x + cos^2x = 1 to simplify the equation further:

25 + 50sinxcosx = 25 + 50(sinxcosx) = 25 + 25sin2x

Now we can see that the left side of the original equation is equal to the right side of the equation, 25sin2x+25.

Therefore, the equation (5sinx+5cosx)^2 = 25sin2x+25 is an identity for all values of x.

Using the intermediate value theorem, the equation x5 - 2x4 - x - 3 = 0 is continuous on the closed interval [2, 3], there is a root of the equation in the interval (2,3).

Given, the equation is x5 - 2x4 - x - 3 = 0

We have to find the root of the equation in the interval (2, 3) using the intermediate value theorem.

Let f(x) = x5 - 2x4 - x - 3

f(2) = (2)5 - 2(2)4 - 2 - 3

f(2) = 32 - 32 - 5

f(2) = -5

f(3) = (3)5 - 2(3)4 - 3 - 3

f(3) = 243 - 162 - 6

f(3) = 75

Intermediate value theorem states that if a continuous function f(x) on the interval [a,b] has values of opposite sign inside an interval, then there must be some value x = c on the interval (a,b) for which f(c) = 0.

f(x) is continuous on the interval [2, 3] because it is a polynomial function, and is continuous at each point in the interval.

Here, f(2) is negative and f(3) is positive.

Therefore, f(x) is continuous on the closed interval [2, 3], there must be some value x = c on the interval [2, 3] for which f(c) = 0.

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The triangles having one angle as 45° have been constructed below.

What is a triangle?

Three vertices make up a triangle, a three-sided polygon. The angles of the triangle are formed by the connection of the three sides end to end at a single point.

We need to draw 2 triangles which contains one angle as 45 degrees.

The first triangle is the triangle with angles 45°, 60° and 75°.

So, the second triangle is a right-angled triangle with angles 90°, 45° and 45°.

The two triangles stated above have been drawn below.

Hence, there are 2 different triangles with one angle as 45°.

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Since there are multiple questions, so the question answered above is as follow:

For the problem, draw 2 triangles that have the listed properties. Try to make them as different as possible.

1. One angle is 45 degrees.

Answer:

f(x) has a global minimum at x = 4.08628

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = x^4 - 4 x^3 - 9 x^2 + x - 16

Find the critical points of f(x):

Compute the critical points of x^4 - 4 x^3 - 9 x^2 + x - 16

To find all critical points, first compute f'(x):

d/(dx) (x^4 - 4 x^3 - 9 x^2 + x - 16) = 4 x^3 - 12 x^2 - 18 x + 1:

f'(x) = 4 x^3 - 12 x^2 - 18 x + 1

Solving 4 x^3 - 12 x^2 - 18 x + 1 = 0 yields x≈-1.13994 or x≈0.0536696 or x≈4.08628:

x = -1.13994, x = 0.0536696, x = 4.08628

f'(x) exists everywhere:

4 x^3 - 12 x^2 - 18 x + 1 exists everywhere

The critical points of x^4 - 4 x^3 - 9 x^2 + x - 16 occur at x = -1.13994, x = 0.0536696 and x = 4.08628:

x = -1.13994, x = 0.0536696, x = 4.08628

The domain of x^4 - 4 x^3 - 9 x^2 + x - 16 is R:

The endpoints of R are x = -∞ and ∞

Evaluate x^4 - 4 x^3 - 9 x^2 + x - 16 at x = -∞, -1.13994, 0.0536696, 4.08628 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-1.13994 | -21.2213

0.0536696 | -15.9729

4.08628 | -156.306

∞ | ∞

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-1.13994 | -21.2213 | neither

0.0536696 | -15.9729 | neither

4.08628 | -156.306 | global min

∞ | ∞ | global max

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-1.13994 | -21.2213 | neither

0.0536696 | -15.9729 | neither

4.08628 | -156.306 | global min

f(x) = x^4 - 4 x^3 - 9 x^2 + x - 16 has one global minimum:

Answer: f(x) has a global minimum at x = 4.08628

:

The the area of the real patio is = 50 m²

Recall that area deals with the floor space occupied by the rectangular field

The area of the lawn is represented by

Area = L*W

2 cm for length = 5 m.

4 cm for width = 10 m.

Area = width • length. 10•5=50cm²

Having computed the area, we now conclude that the the area of the real patio 50cm²

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The total number of candidates in the group will be 50.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that the grade six class went on a field trip. of this group of pupils, only 98% or 49 joined the activity.

The total number will be calculated as,

Total number = [ 49 x 100 ] / 98

Total number = 50

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U1 ≠ U2. The claim is therefore false in general

Let V = R2 over R, U1 the x-axis, U2 the y-axis, and W = V.

Obviously, U1 + W, U2 + W ⊆ V is clear since U1,U2,W ⊆ V .

Let v ∈ V be any vector. Then v ∈ W = V .

Here, instead of vectors, U1, U2, and W are vector spaces. Therefore, we are unable to demonstrate the evidence by using U1, U2, and W as separate vectors.

Here we assume that U1={(x, 0), x ∈ R} and U1={(0, y), y ∈ R} and W=V={(x, y), x,y ∈ R}

So v = 0 + v ∈ U1 + W. Also, v = 0 + v ∈ U2 + W.

Hence, V ⊆ U1 +W, U2 +W.

Thus, U1+W = U2+W = V.

But U1 ≠ U2. The claim is therefore false in general.

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Answer:

her mother is 45 years old

Step-by-step explanation:

daughter = n

mother = n + 22

n + n + 22 = 68

2n = 46

x = 23

n + 22 = 23 + 22 = 45

Answer: mothers age = 45 years old

Step-by-step explanation:

mothers age = 22 + x

daughters age= x

create the equation:

22 + x + x = 68

           2x = 68 - 22

           2x = 46

              x = 23 (daughters age)

in order to get the mother's age you should add 22 to daughter's age (23) as they say the mother is 22 years older than the daughter

therefore: 22 + 23 = 45 years old

To find out how long it will take the hose to fill the larger aquarium, we need to compare the volume of water in each aquarium. The volume of the first aquarium is 8 x 9 x 12 = 864 cubic inches, and the volume of the second aquarium is 22 x 27 x 29 = 14,938 cubic inches. The second aquarium is 14,938 / 864 = 17.2 times larger than the first aquarium.

The time it takes to fill the first aquarium is 5 minutes, so it will take 5 x 17.2 = 86 minutes to fill the second aquarium with the same hose.

It will take 86 minutes to fill the second aquarium.

Option E is the correct option

In this experiment, a farmer completed 4 stages of therapy using the Randomized Design idea.

Complete randomization of the design

In terms of comfort and ease of data processing, a totally random design is essentially the simplest experimental design. Subjects are randomly assigned to treatments in this design.

In fact, randomization is what this fully randomized design depends on to account for the impacts of unimportant factors. In reality, the investigator anticipates that uncontrollable variables will generally impact treatment conditions in an equal manner, so any meaningful variations across conditions may be accurately assigned to the independent variable.

Design for double blinds:

In an experiment, the placebo effect will be diminished or abolished if participants in the control group are aware that they are getting a placebo, and the placebo will no longer function as a control.

Blinding fundamentally involves withholding information about whether or not a person is getting a placebo. Thus, the placebo effect is equally felt by participants in the control and treatment groups. It's common practice to hide the fact that certain groups get placebos from analysts who review the trial. Double blinding is the term for this procedure. It ensures that their appraisal is not affected by knowledge of real treatment circumstances and stops the analysts from "spilling the beans" to the participants through subtle indications.

Designing Matched Pairs .

A randomized block design is a specific example of the matched pairs design. It can be employed when there are essentially just two treatment conditions in an experiment, and participants can be paired off based on some sort of blocking variable. The individuals are then randomly randomized to various treatments within each pair.

Block design using randomization:

In a randomized block design, participants are really split up into smaller groups called blocks so that the variance within blocks is less than the variance between blocks. After that, patients within each block are allocated to treatment scenarios at random.

So, in this experiment, the farmer applied 4 doses of therapy.

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It would take approx 4 years for 5300 to become 7500 if it’s invested at 9% and us compounded quarterly.

The interest earned on a deposit is known as compound interest because it is computed using both the initial principal and the interest accrued over the course of previous periods. Compound interest is simply interest that is earned on interest, to put it another way. Interest can be compounded daily, monthly, or yearly, among other frequency schedules.

Compound interest increases with a higher number of compounding periods. A snowball-like image comes to mind. Your snowball will grow faster if you start saving early and contribute more money to it.

The compound interest formula is given by:

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

Where,

A = final amount

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of time periods elapsed

Here,

A = 7500

P = 5300

r = 9%

n = 4 (quarterly)

t = year to find

Putting the values in the formula

[tex]$7500 = 5300 (1 + \frac{0.09}{4})^{4t}[/tex]

t = 3.90

Thus, It would take approx 4 years for 5300 to become 7500 if it’s invested at 9% and us compounded quarterly.

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You can use collections of objects such as blocks to model a number. Each type of block can be used to show the value of each digit in the number

The values of adjacent places in base-ten representations of numbers are related as each place is always 10 times the value represented by the place to its imme- diate right

Unit blocks, rods, flats, and cubes are the names of base ten block types. The digit in the ones place is represented by the unit blocks. To demonstrate how many tens are in a number, rods are each worth one ten, i.e., ten rods equal one flat, which is worth one hundred each.

Students use base ten blocks, also referred to as multibase arithmetic blocks or Dienes blocks, as a type of math manipulative to learn the basics of addition, subtraction, number sense, place value, and counting.

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Answer:

C) Possible.

Step-by-step explanation:

Hope it helps! =D

Option 1 is the more cost-effective option for preventive maintenance.

Option 1:  Flat fee of $400 per month

Option 2:  Pay a fee of $25 per breakdown

Option 1 is the more cost-effective option since the cost is fixed. This option would provide a guaranteed cost of $400 per month, regardless of the frequency of breakdowns.

On the other hand, Option 2 would cost more in the long run if the frequency of breakdowns is high. The cost would be calculated as follows:

Frequency of Breakdowns x Cost per Breakdown = Total Cost

For example, if the frequency of breakdowns is three times per month, the total cost would be 3 x $25 = $75 per month.

In comparison, the cost of Option 1 would remain at $400 per month, regardless of the frequency of breakdowns. Therefore, Option 1 is the more cost-effective option for preventive maintenance.

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The union of two subspaces of V is a subspace of V if and only if one of the subspaces of v is contained in the other.

Let [tex]$V_1, V_2$[/tex] be two subspaces of V.

Suppose is a subspace of V.

Then if [tex]$x_1 \in V_1$[/tex] and [tex]$x_2 \in V_2$[/tex] then we must have [tex]$x_1 \in V_1 \cup V_2$[/tex] and [tex]$x_2 \in V_1 \cup V_2$[/tex] so that we must have [tex]x_1+x_2 \in V_1 \cup V_2$.[/tex]

But this by definition means [tex]$x_1+x_2 \in V_1$[/tex] or [tex]$x_1+x_2 \in V_2$[/tex].

Only the elements that are already present in the two subspaces are combined.

A subspace is formed by the union of two subspaces of a vector space, and each subspace contains the other.

Either way, by the existence of additive inverses and closure properties for subspaces we have:

[tex]\left(x_1+x_2\right)+\left(-x_1\right) \in V_1[/tex]

Or

[tex]\left(x_1+x_2\right)+\left(-x_2\right) \in V_2[/tex]

By associative/commutative properties we have:

[tex]x_2 \in V_1[/tex] or [tex]x_2 \in V_1[/tex]

Thus we have shown if [tex]$x_1 \in V_1$[/tex] and [tex]$x_2 \in V_2$[/tex] then [tex]$x_1 \in V_2$[/tex] or  [tex]$x_2 \in V_1$[/tex].

If [tex]$x_1 \in V_2$[/tex] for all [tex]$x_1 \in V_1$[/tex] then we have [tex]$V_1 \subseteq V_2$[/tex].

If [tex]$x_2 \in V_1$[/tex] for all [tex]x_2 \in V_2[/tex] then [tex]$V_2 \subseteq V_1$[/tex].

In either case, we see one subspace is a subset of the other.

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The quantity of water that Ms. Morales put in the pool in terms of π would be = =6561π in³

The volume of the pool can be calculated by using the formula of the volume of a cylinder which is = πr²h

Where,

r = 54/2 = 27in

volume = π × 27²× 12

= π × 8748

= 8748π in³

Therefore the quantity of the pool that's as filled with water = 3/4 of 8748π

= 3× 8748/4

= 26244/4

=6561π in³

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The scale factor applied to the triangle is k = 1/2.

When we have a length L and we apply to it a scale factor K, the new length will be:

L' = k*L

Then the scale factor is:

L'/L = k

And if a scale factor is applied to a polygon, like a triangle, all the sides get affected by the scale factor in the same way.

Then if the original length of the given side is:

L = 5ft

And the new length is:

L' = 2.5ft

The scale factor is:

k = 2.5ft/5ft = 1/2

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To make b the subject of the equation, we have: b = 2a + c

Subject of formula is the process expressing a required alphabet in the terms of others in a given expression. It involves the applications of arithmetic operations so as to arrive at the required answer.

Thus, rearranging the equation and make b the subject in the question, we have;

a = 1/2 (b - c)

a = (b - c)/ 2

multiply through by 2

2a = b - c

express b in terms of a and c to have;

b = 2a + c

Therefore, after rearranging the equation, b as a subject is:

b = 2a + c

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Answer:

  65,900 mL

Step-by-step explanation:

You want to know the volume in mL of a sink with a volume of 65.9 dm³.

The relation between dm³ and mL is ...

  1 dm³ = 1000 mL . . . . . . . . . . also, 1 L (liter)

Multiplying the above relation by 65.9, we have ...

  65.9 dm³ = 65,900 mL

__

Additional comment

For many problems involving volumes in liters and dimensions in centimeters or meters, it can be convenient to convert the dimensions to decimeters (dm). Since 1 dm³ = 1 L, that can make the desired volume easily calculated.

a) Both events occur: A ∩ B

b) At least one event occurs: A ∪ B

c) Neither event occurs: A' ∩ B'

d) Exactly one event occurs: (A ∩ B') ∪ (A' ∩ B)

In Event A and B, the intersection A ∩ B tells us the outcome where both events occur. The Union of A and B tells us the outcome where at least one event occurs. The intersection of the complements of A and B (A' ∩ B') tells us the outcome where neither event occurs.

Finally, the union of the intersection of A and the complement of B and the intersection of the complement of A and B ( (A ∩ B') ∪ (A' ∩ B) ) tells us the outcome where exactly one event occurs.

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The approximate number of defective parts the factory produced is 7200.

A percentage is a value per hundredth. Percentages can be converted into decimals and fractions by dividing the percentage value by a hundred.

Given, During a quality control check, a factory found that 5% of the parts it produces are defective.

Let, 'x' be the number of defective parts.

Therefore, 'x' is 5% of 144000 which can be numerically expressed as,

(5/100)×144000.

= 5×1440.

= 7200.

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The ratio between home fires to bush fires is 4:3.

We can determine how much of one quantity is in the other by comparing two amounts of the same units and finding the ratio. Ratios can be divided into two groups. One is the part-to-whole ratio, and the other is the part-to-part ratio. The link between two distinct entities or groupings is depicted by the part-to-part ratio.

Given the number of home fires = 8

number of bushfires = 6

ratio of home fires to bush fires = 8:6

the common factor of 8 and 6 is 2

divide both terms by 2

(8/2) : (6/2)

= 4:3

Hence the rations in the lowest terms is 4:3.

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Answer:

x = 3

Step-by-step explanation:

using the perpendicular bisector theorem B

MB is the perpendicular bisector of JK and JB = KB

then Δ MJK is isosceles with legs being congruent , that is

MJ = MK , so

9x - 18 = 3x ( subtract 3x from both sides )

6x - 18 = 0 ( add 18 to both sides )

6x = 18 ( divide both sides by 6 )

x = 3

The minimum equation initial height at which the car can be released and still stay in contact with the track at the top of the loop is 8.90 m.

hmin = 8.90 m

We can use the equation for centripetal force to solve for the minimum initial height hmin at which the car can be released and still stay in contact with the track at the top of the loop.

Fc = mv2/r

where Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the loop.

We know that the mass of the car is 250 kg, the speed of the car is 14 m/s, and the radius of the loop is 20 m.

We can rearrange the equation to solve for hmin:

Fc * r = m * v2

hmin = (m * v2) / Fc

hmin = (250 kg * (14 m/s)2) / (25000 N)

hmin = 8.90 m

The minimum initial height at which the car can be released and still stay in contact with the track at the top of the loop is 8.90 m.

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The lines represent different routes taken by people walking in a city, where each block is 140m by 140m in size. The routes are determined by randomized variables, such as starting point, destination, and available pathways.

For a route that is 140m by 140m:

1. Calculate the diagonal length of the route: √2 x 140 = 196.4 m

2. Calculate the perimeter of the route: 4 x 140 = 560 m

3. Calculate the area of the route: 140 x 140 = 19600 m2

When calculating a route that is 140m by 140m, the first step is to calculate the diagonal length of the route. This is done by using the Pythagorean Theorem, which is written as √2 x 140. The result of this calculation is 196.4m. The second step is to calculate the perimeter of the route. This is done by multiplying the length of each side of the route by 4, which gives a total of 560m. The final step is to calculate the area of the route. This is done by multiplying the length and width of the route (140m x 140m), which gives a total of 19600m2. By using these calculations, one can easily measure the length, width, and area of a route that is 140m by 140m.

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Answer: Question one:
              a) 6.67 boxes/ hour
              b) 10 boxes/ hour
              c) 3.33 boxes/ hour
              d) 49.93% increase

Step-by-step explanation:

Question two:

a) 33.33 valves/ hour
b) 36.67 valves/ hour
c) 10.02% increase