In a right triangle the legs have lenths of 8 and 15 what is the perimeter of this triangle

Answer:

$125 on the 6th month

Answer:  First - Add your costs to your mark-up. (Costs + Mark-up Amount = Baseline Price), Next - If you want to find the price percentage, (Costs x Mark-up Percentage = Mark-up Amount) (Mark-up Amount + Costs = Baseline Price) Next - Multiply your baseline price by your overhead percentage and figure out what is the amount that you must add to the baseline prices so that the overhead expenses can be covered. (Price x Overhead Percentage = Overhead Contribution)Now you must finally add your overhead contribution to your baseline price: (Overhead Contribution + Baseline Price = Final Price)

In the subtraction problem 17 - 8, the number 9 is the difference, which is the result obtained after subtracting 8 from 17.

In the problem 17 - 8 = 9, the number 9 is the difference. In subtraction, the difference is the result of subtracting one number from another. Here, when we subtract 8 from 17, we get the difference of 9. This is because 17 is the minuend (the number from which another number is subtracted), 8 is the subtrahend (the number that is to be subtracted), and 9 is the difference (the result of the subtraction).

The solution of the expression '' [tex]10\frac{1}{2}[/tex] divided by [tex]2\frac{1}{4}[/tex] '' will be [tex]4 \frac{2}{3}[/tex].

What is Division method?

Division method is used to distributing a group of things into equal parts. Division is just opposite of multiplications.

For example, dividing 20 by 2 means splitting 20 into 2 equal groups of 10.

Given that;

The mathematical expression is;

'' [tex]10\frac{1}{2}[/tex] divided by [tex]2\frac{1}{4}[/tex] ''

Now, Solve the mathematical expression by using division method as;

The mathematical expression is,

[tex]10\frac{1}{2}[/tex] divided by [tex]2\frac{1}{4}[/tex]  =  [tex]10\frac{1}{2}[/tex] ÷ [tex]2\frac{1}{4}[/tex]

Change the mixed fraction into improper fraction as;

⇒ [tex]10\frac{1}{2}[/tex] ÷ [tex]2\frac{1}{4}[/tex]

⇒ [tex]\frac{21}{2}[/tex] ÷ [tex]\frac{9}{4}[/tex]

It can be solve as;

⇒ 21/2 ÷ 9/4

⇒ 21/2 × 4/9

⇒ 14/3

Change into mixed fraction as;

⇒ [tex]\frac{14}{3} = 4 \frac{2}{3}[/tex]

So,  [tex]10\frac{1}{2}[/tex] ÷ [tex]2\frac{1}{4}[/tex] =  [tex]4 \frac{2}{3}[/tex].

Therefore,

The solution of the expression '' [tex]10\frac{1}{2}[/tex] divided by [tex]2\frac{1}{4}[/tex] '' will be [tex]4 \frac{2}{3}[/tex].

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divide 40 by 4/5 =

40*5/4 = 200/4 =50

 Emma can jump 50 inches

We are given the equation:

c = 500 + 1100 x – 100 x^2

Solve for x given the value of c = 2300

Rearrange and substitute:

- 100 x^2 + 1100 x + 500 = 2300

Divide everything by -100:

x^2 – 11 x – 5 = - 23

x^2 – 11 x = - 18

Completing the square:

(x – 11/2)^2 = - 18 + (11/2)^2

x – 11/2 = ±3.5

x = 2, 9

So x is 2 or 9

6 1/2 divided by  1/4

6.5 /0.25 =26 servings

The circle design a class which represents a circle in the (x,y) coordinate plane.

It is described as a set of points, where each point is at the same distance from a fixed point (called the centre of a circle)

from math import dist # to calculate distance between two points

# defining class circle

class Circle:

   # constructor of Circle

   # with default parameters

   def __init__(self, x_coord=0, y_coord=0, radius=1):

       self.centre_x = x_coord

       self.centre_y = y_coord

       self.radius = radius

   # getters

   def getX(self):

       return self.centre_x

   def getY(self):

       return self.centre_y

   def getRadius(self):

       return self.radius

   # setters

   def setX(self, x):

       self.centre_x = x

   def sety(self, y):

       self.centre_y = y

   def setRadius(self, r):

       self.radius = r

   # function to calculate and return area

   def getArea(self):

       return 3.141*(self.radius**2)

   # function to calculate and return perimeter

   def getPerimeter(self):

       return 2*3.141*self.radius

   # calculates the distance between circle centre and

   # given coordinates

   # returns true if distance is less than radius

   def isPointWithinCircle(self, x, y):

       return dist([self.centre_x, self.centre_y], [x, y]) < self.radius

c1 = Circle(3, 3, 7)

c2 = Circle()

print("First Circle perimeter: " + str(c1.getPerimeter()))

print("First Circle area: " + str(c1.getArea()))

c2.setX(3)

c2.sety(3)

c2.setRadius(2)

if c2.isPointWithinCircle(4, 3.7):

   print("(4, 3.7) is within circle two")

else:

   print("(4, 3.7) is not within circle two")

print("Moving second Circle")

c2.setX(3+c2.getX())

if c2.isPointWithinCircle(4, 3.7):

   print("(4, 3.7) is within circle two")

else:

   print("(4, 3.7) is not within circle two")

Thus, the circle design a class which represents a circle in the (x,y) coordinate plane.

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Answer:

The probability of a positive test result is 0.017919

Option C is correct.

The probability is quite​ low, indicating that further testing of the individual samples will be a rarely necessary event.

Step-by-step explanation:

Probability of finding bacteria in one public swimming pool = 0.009

We now require the probability of finding bacteria in the combined test of two swimming pools. This probability is a sum of probabilities.

Let the two public swimming pools be A and B respectively.

- It is possible for public swimming pool A to have bacteria and public swimming pool B not to have bacteria. We would obtain a positive result from testing a mixed sample of both public swimming pools.

- It is also possible for public swimming pool A to not have bacteria and public swimming pool B to have bacteria. We would also obtain a positive result from testing a mixed sample of both public swimming pools.

- And lastly, it is possible that both swimming pools both have bacteria in them. We will definitely get a positive result from this too.

So, if P(A) is the probability of the event of bacteria existing in public swimming pool A

And P(B) is the probability of the event of bacteria existing in public swimming pool B

P(A') and P(B') represent the probabilities of bacteria being absent in public swimming pool A and public swimming pool B respectively.

P(A) = P(B) = 0.009

P(A') = P(B') = 1 - 0.009 = 0.991

Since the probabilities for each public swimming pool is independent of the other.

P(A or B) = P(A n B') + P(A' n B) + P(A n B)

= P(A)×P(B') + P(A')×P(B) + P(A)×P(B)

= (0.009×0.991) + (0.991×0.009) + (0.009×0.009)

= 0.008919 + 0.008919 + 0.000081

= 0.017919

Evidently, a probability of 0.017919 (1.7919%) indicates an event with a very low likelihood. A positive result is expected only 1.7919% of the time.

Hence, we can conclude that the probability is quite​ low, indicating that further testing of the individual samples will be a rarely necessary event.

Hope this Helps!!!

If Bryson purchased 40 raffle tickets. There were 500 raffle tickets sold. Then 2/25 is the theoretical probability of Bryson winning the trip

It is a branch of mathematics that deals with the occurrence of a random event.

Given,

Bryson hopes to win a three-day vacation in a drawing that is being held at his office.

Bryson purchased 40 raffle tickets

The number of raffle tickets sold were 500

We need to find the theoretical probability of Bryson winning the trip

P(win)=40/500

=4/50

=2/25

2/25 because he bought 40 out of the 500 tickets

Hence 2/25  is the theoretical probability of Bryson winning the trip

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To solve such problems we must know about Tangent (Tanθ).

The tangent or tanθ in a right angle triangle is the ratio of its perpendicular to its base. it is given as,

[tex]\rm{Tangent(\theta) = \dfrac{Perpendicular}{Base}[/tex]

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The base is the adjacent smaller side of the angle θ.

The height of the mountain is 1.495 miles.

Given to us

[tex]\rm{ tan(\angle A) = \dfrac{Perpendicular}{Base}[/tex]

[tex]tan(\angle A) = \dfrac{CD}{AD}[/tex]

[tex]tan(3.5^o) = \dfrac{CD}{15+x}[/tex]

[tex]CD=tan(3.5^o) \times(x+15)[/tex]

[tex]\rm{ tan(\angle B) = \dfrac{Perpendicular}{Base}[/tex]

[tex]tan(\angle B) = \dfrac{CD}{BD}[/tex]

[tex]tan(9^o) = \dfrac{CD}{x}[/tex]

[tex]CD=tan(9^o) \times(x)[/tex]

Equating CD,

[tex]tan(3.5^o) \times (x+15) = tan(9^o) \times x\\\\(x+15)= \dfrac{tan(9^o) }{tan(3.5^o) }\times x\\\\x+15=2.59\ x\\15 = 2.59x-x\\15= 1.59x\\x = 9.44[/tex]

Substituting the value of x

[tex]CD=tan(9^o) \times(x)[/tex]

[tex]CD=tan(9^o) \times(9.44)\\CD= 1.495[/tex]

Hence, the height of the mountain is 1.495 miles.

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The fraction which is halfway between  [tex]\frac{1}{2}[/tex] and [tex]\frac{5}{6}[/tex] is equal to [tex]\frac{2}{3}[/tex].

What is a fraction ?

A fraction is used to represent a portion or component of a whole. The numerator and denominator are its two parts.

The two fractions are given which are [tex]\frac{1}{2}[/tex] and [tex]\frac{5}{6}[/tex].

We know that numerator is the top value of the fraction whereas the denominator is the bottom value of the fraction. We need to add the given fractions and then divide it by 2 to find the halfway between the two given functions.

To add the fractions we need to equal denominator firstly. The common denominator of 2 and 6 will be 6 because 6 is a multiple of both 2 and 6.

So ,

Halfway between  [tex]\frac{1}{2}[/tex] and [tex]\frac{5}{6}[/tex] is :

= ( [tex]\frac{1}{2}[/tex] + [tex]\frac{5}{6}[/tex] ) / 2

= [tex]\frac{3 + 5}{6}[/tex] / 2

= [tex]\frac{8}{6}[/tex] × [tex]\frac{1}{2}[/tex]

= [tex]\frac{4}{6}[/tex]

or

[tex]\frac{2}{3}[/tex]

Therefore , the fraction which is halfway between  [tex]\frac{1}{2}[/tex] and [tex]\frac{5}{6}[/tex] is equal to [tex]\frac{2}{3}[/tex].

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After walking 12.5 miles with a goal of 20 miles, you need to walk 7.5 more miles to meet your goal.

To meet your goal in the charity walk, you need to calculate how many more miles are left to walk after having walked 12.5 miles with a goal of 20 miles. The calculation is simple subtraction: 20 miles (goal) - 12.5 miles (distance already walked) = 7.5 miles remaining to be walked. So, you need to walk 7.5 more miles to meet your goal.

Answer:  The length of the chord AB is √2 r units.

Step-by-step explanation: Given in the question that Circle P is tangent to the X-axis at the point A and tangent to the Y-axis at the point B.

Also, the co-ordinates of centre of the circle P are C(r, r). See the attached image below.

We are to find the length of the chord AB, with end-points A and B as the points of tangency.

From the figure, we note that

the co-ordinates of the points A and B are  (0, r) and (r, 0) respectively.

As the X-axis and Y-axis are perpendicular to each other, so the lines parallel to them, OB and OA are also perpendicular to each other.

That is, ∠ACB = 90°.

This implies that the triangle ACB is a right-angled triangle.

Now, using Pythagoras theorem in ΔACB, we have

[tex]AB^2=CA^2+CB^2~~~~~~~~~~~~~~~~~~~~~(i).[/tex]

The lengths of the line segments CA and CB are calculated, using distance formula, as follows:

[tex]CA=\sqrt{(0-r)^2+(r-r)^2}=\sqrt{r^2}=r,\\\\CB=\sqrt{(r-r)^2+(0-r)^2}=\sqrt{r^2}=r.[/tex]

From equation (i), we get

[tex]AB^2=r^2+r^2\\\\\Rightrarrow AB^2=2r^2\\\\\Rightarrow AB=\sqrt2r~\textup{units}.[/tex]

Thus, the length of the chord AB is √2 r units.

A circle, tangent to the x-axis and y-axis, forms a square where its radius r is the side length. The chord, which is the diagonal, can be found using the Pythagorean theorem resulting to r√2.

In the given problem, a circle P is tangent to the x-axis and y-axis at two points. This means that the radius of the circle is the distance from the center to either axis, which are both r, as the coordinates of the center are (r,r). We are asked to find the length of the chord whose endpoints are the points of tangency.

The chord is actually the diagonal of a square that has sides of length r. So, its length can be found using the Pythagorean theorem (a^2 + b^2 = c^2), where a and b are the lengths of the sides of the square and c is the diagonal. Hence the length of the chord can be found as √(r^2 + r^2) = √(2r^2) = r√2.

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