Use the scale of 1cm=45 km
The actual distance between two cities is 333 km. What is the distance between them on the map

Answer:

200 meters

Step-by-step explanation:

If the ratio is 5 inches to 20 meters that means for every 5 inches in the model there will be 20 meters on the real ship.

If the model is 50 inches long that means that there are 10 5 inch segments, multiply this 10 by the 20 meters in the ratio and you will get 200 meters as the final length for the ship.

Answer:

x = 4

Step-by-step explanation:

Since M is the midpoint of segment KL, then segments KM and LM are congruent. They have the same length.

4x + 1 = 8x - 15

4x = -16

x = 4

The graph depicts that at the end of the snowstorm, there was 5 inches of snow left. See the attached graph.

It is not be noted that the total amount of hours which the snow fell is:

2 + 9 = 11 hours.

Although the graph depicts a peak of 14 hours, recall that there were three hours when the snow didn't fall.

While the maximum depth of the snow was 11 inches.

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Applying the angle of intersecting secants theorem, the measure of angle B is: 35°.

The measure of the angle formed outside a circle by two intersecting secants is equal to half the positive difference of the measures of the arcs they intercept based on the angle of intersecting secants theorem.

Applying the angle of intersecting secants theorem, we have the following:

Let the missing measure of the arc for angle A be x. Therefore:

6 = 1/2(x - 17)

2(6) = x - 17

12 = x - 17

12 + 17 = x

x = 29°

m∠B = 1/2(99 - x)

Plug in the value of x

m∠B = 1/2(99 - 29)

m∠B = 35°

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Using proportions, it is found that their bill before adding the tip was of $27.48.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

They paid a 20.2% tip, hence 120.2% = 1.202 of the price x is equals to $33.03, hence the bill before the tip is found as follows:

1.202x = 33.03

x = 33.03/1.202

x = $27.48.

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Answer:

#3

Step-by-step explanation:

To create a number line to represent x<= 0, draw a solid circle at 0 (because 0 is included), then draw an arrow extending to the left (because it's less than or equal to).

A $2.00 base fare charge, and per-mile, and per-minute rates related as follows; 3•x + 7•y = 6.5 and 7•x + 14•y = 14, give the distance traveled in the third ride as 4.5 miles

The base fare = $2.00

Let x represent the per-mile rate, and let y represent the per-minute rate, we have;

The cost of Ryan's first taxi ride = $8.50

Distance traveled in the first ride = 3.0 miles

Time taken during the first ride = 7 minutes

Therefore;

2 + 3•x + 7•y = 8.5

Which gives;

3•x + 7•y = 8.5 - 2 = 6.5

Distance traveled in the second ride = 7.0 miles

Duration of the second ride = 14 minutes

The second ride cost = $16.00

Therefore;

2 + 7•x + 14•y = 16

Which gives;

7•x + 14•y = 16 - 2 = 14

Solving the above simultaneous equations by multiplying equation (1) by 2 then subtracting the result from equation (2) gives;

(7•x + 14•y) - 2 × (3•x + 7•y) = 14 - 2×6.5 = 1

x = 1

3•x + 7•y = 6.5

7•y = 6.5 - 3•x

7•y = 6.5 - 3×1 = 3.5

y = 3.5/7 = 1/2 = 0.5

Duration of the third taxi ride = 10 minutes

Cost of the third ride = $13.50

Therefore;

2 + 1×a + 10×y = $13.50

Where a is the distance traveled during the third ride, we have;

2 + 1×a + 10×0.5 = $13.50

2 + a + 5 = 13.5

a = 13.5 - 2 - 7 = 4.5

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If the radius of the circular table is 5 cm then the length of the trim required is 31.4 cm.

Given that the radius of the circular table is 5 cm.

What is the length of trim needed to decorate along the outside trim?

Circumference is the length of arc of the circle.It is also known as the perimeter of the  circle.

Circumference of the circle=2πr in which r is the radius of the circle.

It is given that the trim is placed and decorated along the outside edge, so the perimeter of the circle must equal to the length of trim needed.

Length of trim needed to decorate the circular table=2πr

=2*π*5

=10π

=10*3.14

=31.4 cm.

Hence the length of the trim needed to decorate along the table is 31.4 cm.

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Answer:

Step-by-step explanation:

Let the fraction be x/y.

According to question we have the following equations.

The denominator of a fraction exceeds numerator by 3:

If the numerator is doubled and the denominator is increased by 14, then fraction becomes 2/3rd of the original fraction:

Change the fraction as below and solve for y:

Find the value of x using the first equation:

The fraction is:

Answer:

Original fraction = ⁴/₇

Step-by-step explanation:

Numerator:  top of the fraction

Denominator: bottom of a fraction

Let x be the original numerator.

If the denominator of a fraction exceeds the numerator by 3:

[tex]\implies \dfrac{x}{x+3}[/tex]

If the numerator is doubled and the denominator is increased by 14, then fraction becomes 2/3rd of the original fraction:

[tex]\implies \dfrac{2x}{x+3+14}=\dfrac{2}{3}\left(\dfrac{x}{x+3}\right)[/tex]

[tex]\implies \dfrac{2x}{x+17}=\dfrac{2x}{3(x+3)}[/tex]

[tex]\implies \dfrac{2x}{x+17}=\dfrac{2x}{3x+9}[/tex]

Cross multiply:

[tex]\implies 2x(3x+9)=2x(x+17)[/tex]

Divide both sides by 2x:

[tex]\implies 3x+9=x+17[/tex]

Subtract x from both sides:

[tex]\implies 2x+9=17[/tex]

Subtract 9 from both sides:

[tex]\implies 2x=8[/tex]

Divide both sides by 2:

[tex]\implies x=4[/tex]

Substitute the found value of x into the original fraction:

[tex]\implies \dfrac{4}{4+3}=\dfrac{4}{7}[/tex]

Therefore, the original fraction is ⁴/₇.

Step 1. You need x isolated.
[tex]x < 13+7[/tex]

When you move a number from the left to right (and its addition or subtraction, the sign stays (< ≤ > and ≥ is signs).
x<20 or in words, x is less than 20, and cannot be 20.

Answer:

x < 20

Any number less than 20 is a solution

Step-by-step explanation:

x-7 < 13

The first step to solve this inequality is to add 7 to each side

x-7+7 < 13+7

x < 20

Any number less than 20 is a solution

The relative frequency of students who have three siblings is 25% given that there are 60 students in a class in which 5 have no siblings, 26 have one sibling, 14 have two siblings and 15 have three siblings. This can be obtained by using the formula for relative frequency.

Given that,

total number of students in the class = 60

number of students who have no sibling = 5

number of students who have 1 sibling = 26

number of students who have 2 sibling = 14

number of students who have 3 sibling = 15

Formula for relative frequency = f/n, where f is the number of times the data occurred, n is the total number of frequencies.

Therefore,

relative frequency of students who have three siblings = 15/60 = 0.25

In percentage ⇒ 0.25 × 100 = 25%

Hence the relative frequency of students who have three siblings is 25% given that there are 60 students in a class in which 5 have no siblings, 26 have one sibling, 14 have two siblings and 15 have three siblings.

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The 95% confidence interval for the standard deviation of the height of students at UH is given by; CI = (1.81, 4.03)

The formula for the confidence interval for the standard deviation is given by the formula;

CI = √[(n - 1)s²/(χ²ₙ ₋ ₁, α/2)], √[(n - 1)s²/(χ²ₙ ₋ ₁, (1 - α)/2)]

We are given;

Sample size; n = 14

D F = n - 1 = 14 - 1 = 13

Standard Deviation; s = 2.5

Confidence Level; CL = 95% = 0.95

Significance level; α = 1 - 0.95 = 0.05

Thus, using Chi-square distribution table online we have;

χ²₁₃, ₀.₀₂₅  = 24.736

(χ²₁₃, ₀.₉₇₅) = 5.01

Now, the 95% confidence interval for the standard deviation of the height of students at UH is given by :-

CI = √[(13 * 2.5²/(24.736)], √[(13 * 2.5²/(5.01)]

CI = (1.81, 4.03)

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The correct probability that Juan's bus is going to be late every week next week is 20 percent.

We have the total number in the stimulation on to be from 0 to 9

On the fact that it would be late, the number ranges from 2 to 9

Hence the fact that it would be late would be

2/10

= 0.2

0.2 is also the same as 20 percent.

Juan rides the bus to school each day. He always arrives at his bus stop on time, but his bus is late 80% of the time. Juan runs a simulation to model this using a random number generator. He assigns these digits to the possible outcomes for each day of the week:

• Let 0 and 1 = bus is on time

• Let 2, 3, 4, 5, 6, 7, 8, and 9 = bus is late

The table shows the results of the simulation.

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Answer:

x = 1/3 or x = -2/3

Step-by-step explanation:

Let's solve your equation step-by-step.

9x2+3x−2=0

For this equation: a=9, b=3, c=-2

9x2+3x+−2=0

Step 1: Use quadratic formula with a=9, b=3, c=-2.

[tex]x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]x=\frac{-(3)\pm\sqrt{(3)^2-4(9)(-2)} }{2(9)}[/tex]

[tex]x=\frac{-3\pm\sqrt{81} }{18}[/tex]

x = 1/3 or x = -2/3

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{What is the quadratic formula?}[/tex]

[tex]\rm{x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}[/tex]

[tex]\textsf{Or even}[/tex]

[tex]\rm{ax^2 + bx + c = 0}[/tex]

[tex]\huge\textbf{What are we looking for?}[/tex]

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

[tex]\huge\textbf{What are the labels in the equation?}[/tex]

[tex]\mathsf{a \rightarrow 9}\\\\\mathsf{b\rightarrow 3}\\\\\mathsf{c \rightarrow -2}[/tex]

[tex]\huge\textbf{Solving for your equation:}[/tex]

[tex]\rm{x = \dfrac{-(3)\pm \sqrt{3^2 - 4(9)(-2)}}{2a}}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\rm{x = \dfrac{-3\pm \sqrt{81}}{18}}[/tex]

[tex]\rm{x = \dfrac{1}{3}\ or\ x = - \dfrac{2}{3}}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\frak{\mathsf{x} = \dfrac{1}{3}\ or\ \mathsf{x} = - \dfrac{2}{3}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

30%   discount

Step-by-step explanation:

(35 - 24.50) / 35  * 100% = 30%

There were 74 problems in the test.

Let the student got total number of problems in the test to be X.

Percentage of correct answer = 39 %

⇒ 61% of X were incorrect

⇒61/100 x X = 45

⇒61X = 45x100

⇒X = 4500/61

⇒X=73.77

After rounding off 73.77, we get 74.

Hence, we can say that there were total of 74 problems in the test.

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Answer:

It's attached you may see it!!

Answer:

see attachments

Step-by-step explanation:

Given equations:

[tex]\begin{cases}5x + 7y = 50\\7x + 5y = 46\end{cases}[/tex]

To plot the graphs of the given equations:

Equation 1

[tex]\implies 5x + 7y = 50[/tex]

[tex]\implies 7y = -5x + 50[/tex]

[tex]\implies y = -\dfrac{5}{7}x+\dfrac{50}{7}[/tex]

[tex]x=-4 \implies y = -\dfrac{5}{7}(-4)+\dfrac{50}{7}=10 \implies (-4,10)[/tex]

[tex]x=3 \implies y = -\dfrac{5}{7}(3)+\dfrac{50}{7}=5 \implies (3,5)[/tex]

Plot the points (-4, 10) and (3, 5) then draw a straight line through them (see attachment 1).

Equation 2

[tex]\implies 7x+5y=46[/tex]

[tex]\implies 5y=-7x+46[/tex]

[tex]\implies y=-\dfrac{7}{5}x+\dfrac{46}{5}[/tex]

[tex]x=3 \implies y=-\dfrac{7}{5}(3)+\dfrac{46}{5}=5 \implies (3,5)[/tex]

[tex]x=8 \implies y=-\dfrac{7}{5}(8)+\dfrac{46}{5}=-2 \implies (8,-2)[/tex]

Plot the points (3, 5) and (8, -2) then draw a straight line through them (see attachment 2).

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Answer:

The circumference

Step-by-step explanation:

[tex]circumference \: = 2\pi \times radius \\ c = 2\pi(4.4) \\ c = 8.8\pi = 27.65[/tex]

Answer:

555 m

Step-by-step explanation:

Because the model should have identical proportions to the real-life tower, the ratio between the height and shadow length of both diagrams should be the same.

Height (model)             Height (real)
-------------------------  =  ------------------------
Shadow (model)         Shadow (real)

Math:

 0.9 meters                   x
--------------------  =  ----------------------               <----- Ratio
 0.2 meters           123.3 meters

4.5 = x / 123.3 meters                                  <----- Divide 0.2 from 0.9

555 meters = x                                            <----- Multiply both sides by 123.3

The perimeter of the quadrilateral illustrated is 30cm.

The information is incomplete. Therefore, an overview will be given. It should be noted ha perimeter simply means the addition of the sides given.

Let's assume that that we've a rectangle with length and width of 10 and 5cm.

The perimeter will be:

= 2(l + w)

= 2(10 + 5)

= 30cm

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There is a 9/25 chance that two numbers will be added together such that their total exceeds their product.

The set of positive numbers are:

{1,2,3,4,5}

We have to independently select two numbers from these sets.

We are trying to determine the likelihood that the sum is larger than the product of the integers.

When one of the chosen numbers is 1, the total will always be greater than the product because:

1*1=1 & 1+1=2

1*2=2 & 1+2=3

1*3=3 & 1+3=4

1*4=4 & 1+4=5

1*5=5 & 1+5=6

and so on.

If we select 2 as both numbers then,

2*2=4 & 2+2=4

Here sum and the product are equal.

If not, the product will be bigger than the sum.

Now, the first stage is to evaluate the total number of two combinations that are feasible:

We have 5 alternatives for the first number and 5 alternatives for the second number.

The product of the number of possibilities in each scenario yields the total number of combinations:

C= 5*5=25

The following combinations have a sum that is greater than the product :

1 and 1

1 and 2

1 and 3

1 and 4

1 and 5

2 and 1

3 and 1

4 and 1

5 and 1

So, we get 9 combinations.

So, Probability,p=9/25

Therefore it is concluded that the probability is 9/25.

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The value of positive integers in the set are 16.

According to the statement

we have given that the there is a set of numbers from 1 to n and we have to find that the how many integers in this set. and there is one condition that the numbers in the set are less than or equal to 24.

So, For this purpose,

Since [tex]$1 + 2 + \cdots + n = \frac{n(n+1)}{2}$[/tex]

the condition is equivalent to having an integer value for [tex]$\frac{n!} {\frac{n(n+1)}{2}}$.[/tex]

This reduces, when [tex]$n\ge 1$[/tex], to having an integer value for [tex]$\frac{2(n-1)!}{n+1}$[/tex]

This fraction is an integer unless n+1 is an odd prime. There are 8 odd primes less than or equal to 24,

so there are 24-8 = 16.

So, The value of positive integers in the set are 16.

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Vic and Dan are 2, 897m apart.

It is important to note that the distance between Vic and Dan is the base of CN

Let's say the distance to Dan is x

The distance to Vic is y

Using cosine ratio, we have

cos α = opposite / adjacent

α = 72°

opposite = 553. 3cm

Adjacent = x

cos  72° = [tex]\frac{553. 3}{x}[/tex]

Cross multiply

[tex]cos 72[/tex] × [tex]x[/tex] = [tex]553. 3[/tex]

[tex]0. 3090x= 553. 3[/tex]

[tex]x = \frac{553. 3}{0. 3090}[/tex]

[tex]x = 1, 790. 61[/tex] m

The distance to Vic is y

Using the cosine ratio, we have

[tex]cos 60 = \frac{553. 3}{y}[/tex]

Cross multiply

[tex]0. 5y = 553. 3[/tex]

[tex]y = \frac{553. 3}{0. 5}[/tex]

[tex]y = 1,106. 6[/tex]m

To determine how far apart Vic and Dan, we use = x + y

= 1790. 61 + 1106. 6

= 2, 897. 21m

= 2, 897m

Thus, Vic and Dan are 2, 897m apart.

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See below for the solution to each question

Yes, the graph is a function.

This is because all x values have different y values

This is the set of input values of the graph.

From the graph, we have

x = 0 to x = 17

Hence, the domain is [0, 17]

This is the set of output values of the graph.

From the graph, we have

y = 0 to y = 10

Hence, the range is [0, 10]

This is the maximum point on the graph.

From the graph, we have

Maximum = (12, 10)

This is the minimum point on the graph.

From the graph, we have

Minimum = (0, 0)

These are the intervals where the y values increase as x increase.

From the graph, we have

Increasing intervals = (0, 5) ∪ (10, 12) ∪ (14, 15)

These are the intervals where the y values decrease as x increase.

From the graph, we have

Decreasing intervals = (7, 10) ∪ (12, 14) ∪ (15, 17)

These are the intervals where the y values remain unchanged as x changes.

From the graph, we have

Constant intervals = (5, 7)

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Answer:

The estimated result is 680, which suggests the calculator is incorrect.

answer choices would have helped, hope this helps :)

The volume of cone whose area of base is 8π[tex]mm^{2}[/tex] is 32π/3 [tex]mm^{3}[/tex].

Given area of base of the cone 8π[tex]mm^{2}[/tex].

We are required to find  the volume of the cone.

We know that base of a cone used to be in circle  so the area of base of cone is equal to π[tex]r^{2}[/tex].

Area=8π   (given)

π[tex]r^{2}[/tex]=8π

[tex]r^{2}[/tex]=8

r=[tex]\sqrt{8}[/tex]

r=2[tex]\sqrt{2}[/tex]

Volume of cone=1/3*π[tex]r^{2} h[/tex]

=1/3*π[tex](2\sqrt{2} )^{2}[/tex]*4

=1/3*8*4π

=32π/3

(We are not required to put value of π so our answer will be 32π/3.)

Hence the volume of cone whose area of base is 8π is 32π/3.

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Question is incomplete as it should also include height of cone be 4 mm.

Answer:

11/20

Step-by-step explanation:

1/4= 10/40

3/5=24/40

3/10=12/40

10/40+24/40= 34/40

34/40 - 12/40= 22/40

simplifies to 11/20