What were the two whole numbers?

m=3/2 is the answer for the equation

It will take 16 tiles to cover 1 square yard.

To find out how many tiles it will take to cover 1 square yard, we need to calculate the area of one tile.

The tiles are 1/4 by 1/4 yard, so the area of one tile is (1/4) * (1/4) = 1/16 square yard.

To cover 1 square yard, we need to divide the total area by the area of one tile:

1 square yard / (1/16 square yard) = 1 * (16/1) = 16 tiles.

For this case, we have the following polynomial:

[tex]3x ^ 6yz + x ^ 4y ^ 2 + 2xz ^ 5-6x ^ 2y ^ 3-7y\\[/tex]

A term of the polynomial composed by:

[tex]3x ^ 6yz[/tex], where:

3: Represents the coefficient of the term

x, y, z: Represent the associated variables

6: Represents the exponent associated with the variable x. (y, z have exponent 1)

Thus, using the previous definition, it is observed that the expression has 5 terms.

Answer:

Option c

Answer:

Please find the attachment.

Step-by-step explanation:

We are asked to graph the equation [tex]y-5=-\frac{2}{3}(x+9)[/tex].

We know that point-slope form of an equation is in form [tex]y-y_1=m(x-x_1)[/tex], where, m represents slope of line and [tex](x_1,y_1)[/tex] is a point on the line.

We can see that slope of the line is [tex]-\frac{2}{3}[/tex], so line will be a decreasing line.

We can represent the given equation as [tex]y-5=-\frac{2}{3}(x-(-9))[/tex].

The line will pass through the point [tex](-9,5)[/tex].  

The graph of the linear equation can be seen in the image at the end.

To graph any line, we just need to identify two points in the line, and then we can connect them with a line.

Here the equation is:

y - 5 = (-2/3)*(x + 9)

We can see that this equation contains the point (-9, 5) (both sides of the equation become zero in that point)

5 - 5 = (-2/3)*(-9 + 9)

0 = 0

And evaluating in x = 0 we get:

y - 5 = (-2/3)*(0 + 9)

y = 5 + (-2/3)*9

y = 5 - 6

y = -1

So we have the point (0, -1)

Just graphing these two points and conecting them with a line we will get the following graph:

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Final answer:

The conditional probability formula for event B given event A is represented as P(B|A), which is calculated using P(A and B) / P(A). Using dice as an example, the events are described and demonstrated to be neither mutually exclusive nor independent due to overlapping outcomes and interdependence.

Explanation:

The conditional probability of event B given that event A has occurred is represented as P(B|A). To fill in the blanks in the provided formula, it would be P(B | A) = P(B ∩ A) / P(A). Using the provided scenarios, we need to calculate P(A), which is the probability that the sum of numbers on both cubes is less than 10, P(B), which is the probability that the sum of numbers on both cubes is a multiple of 3, and P(A and B), which is the probability that both events occur simultaneously. Finally, we use the formula P(A|B) = P(A and B) / P(B) to calculate the conditional probability.

For example, event A could be rolling a 3 or 4 first, followed by any even number, and event B could be rolling a sum at most 7 on two dice. These events are not mutually exclusive since they can both occur simultaneously, nor are they independent since the occurrence of one affects the likelihood of the other happening.

The completed conditional-probability formula is: P(B|A) = P(A ∩ B) / P(A)

To complete the conditional-probability formula for event B given that event A occurs first, we need to fill in the blanks with the appropriate terms:

P(B|A) = P(A ∩ B) / P(A)

Let's break down each term:

a0: Event A occurs

a1: Event B occurs

a2: Event A and B occur simultaneously (intersection of A and B)

a3: Event A occurs

a4: Event B occurs

Therefore, the completed conditional-probability formula is:

P(B|A) = P(A ∩ B) / P(A)

The answer is 111 lol

1 multiplied by anything is the given number

so 111 x 1 = 111

Answer:

E

Step-by-step explanation:

Construction of a perpendicular to a line through a given point on the line might this image result from. Correct option is C.

A. Construction to copy a line segment: This construction involves using a straightedge and compass to create an identical copy of a given line segment.

B. Construction of a perpendicular bisector: This construction involves finding the perpendicular bisector of a given line segment, which is a line that intersects the segment at its midpoint and forms right angles with it.

C. Construction of a perpendicular to a line through a given point on the line: This construction involves drawing a line perpendicular to a given line through a specific point on that line.

D. Construction of a line parallel to an existing line: This construction involves drawing a line that is parallel to a given line and passes through a specific point not on the given line.

E. Construction of a perpendicular to a line through a given external point: This construction involves drawing a line perpendicular to a given line and passing through a point located outside the given line.

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i am pretty sure it is mADC

The value of m is -2 for 9 = -7m + 1 - 6.

Given Equation;

9 = -7m + 1 - 6

Add 6 on both side,

9 + 6 = -7m + 1 - 6 +6

15 = -7m + 1

Subtract 1 on both side,

15 -1 = -7m + 1 -1

14 = -7m

m = - 2

Therefore, the value of m is -2.

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Answer:

The value of the (RIC) will increase from 4 to 5.75, that is, 44%

Step-by-step explanation:

To answer this question you have to know the definition of Rank well.

The range is defined as the difference between the maximum and minimum value of a series of data. Xmax - Xmin

The interleaving range (RIC) is a measure of dispersion that measures the central range of 50% of the data.

Therefore, if a low value is included, such as five, the variance of the data would be greater and, consequently, the value of the (RIC) will increase from 4 to 5.75, that is, 44%

Answer:

Hence the interquartile range increased by 3(7-4) when we included the eight weight.

Step-by-step explanation:

The weight of the seven rocks is given as:

11     13     14     6     10     9      10

on arranging the data in the ascending order we get the observation as:

6     9    10    10    11     13    14

we divide our data into 3 sets:

the median of data([tex]Q_2[/tex]) is: 10 (as it is the middle value among the data)

The lower set of data is:

6    9    10

[tex]Q_1=9[/tex] ( as it is the middle value)

The upper set of data is:

11     13     14

[tex]Q_3=13[/tex]  ( as it is the middle value in the upper set of data)

Hence, the interquartile range is: [tex]Q_3-Q_1=13-9=4[/tex]

The weight of eight rocks is given as:

11      13     14      6      10     9    10     5

On arranging the data in ascending order we get:

5       6        9     10     10    11      13     14

The median of the data is denoted by [tex]Q_2[/tex] which is the middle value of the given data.

Hence the median here is between 10, 10.

so, the median is given by:

[tex]\dfrac{10+10}{2}=10[/tex]

thus [tex]Q_2=10[/tex]

also the lower set of data is:

5    6    9

Thus [tex]Q_1=6[/tex] (as it is the middle value in the lower set of data)

the upper set of data is:

11   13   14

Thus [tex]Q_3=13[/tex]

Hence, the interquartile range is:

[tex]Q_3-Q_1=13-6=7[/tex]

So when we were having seven weights the interquartile range was: 4

and when we included the eight weight the interquartile range becomes: 7

Hence the interquartile range increased by 3(7-4) when we included the eight weight.

D

given the equation of a parabola in standard form : ax² + bx + c = 0 ( a ≠ 0 )

then the x-coordinate of the vertex  = - [tex]\frac{b}{2a}[/tex]

y = - 2x² + 4x + 12 is in standard form

with a = - 2, b= 4 and c = 12

[tex]x_{vertex}[/tex] = - [tex]\frac{4}{-4}[/tex] = 1

substitute x = 1 into the equation for y

y = - 2(1)² + 4(1) + 12 = 14

vertex = (1, 14 )

Answer: the answer is d

Let x the number of brochures and y the number of fliers.

Optimizing function⇒ f(x,y) = 0.1 x + 0.06 y

Constraints:

x ≥ 80

y ≥ 100

3x + 2y ≤ 500

See the attached figure which represents the constraints.

From the attached figure:

The vertices of the feasible region are (80, 100), (80, 130) and (100, 100).

Substitute the points (80, 100), (80, 130) and (100, 100). in the function

f(x,y) = 0.1 x + 0.06 y

If (x,y) = (80, 100) ⇒ f(x,y) = 14

If (x,y) = (80, 130) ⇒ f(x,y) = 15.8

If (x,y) = (100, 100) ⇒ f(x,y) = 16

The minimum value 14 at (80, 100).

To minimize the cost, she would print 80 brochures and 100 fliers.

This is an optimization problem in mathematics that can be solved by linear programming. The constraints of the problem can be represented by inequalities, and the objective function (cost of printing) can be minimized by finding the optimal number of brochures and fliers to print.

This question is a type of optimization problem in mathematics, specifically a type of problem often found in linear programming. The objective is to minimize the cost of printing brochures and fliers.

First, set up the inequalities based on the constraints. Let x be the number of brochures and y be the number of fliers.

Then, we need to find the number of brochures and fliers to minimize the cost, which is 0.10x + 0.06y. You can solve this problem using methods of linear programming, such as graphing the feasible region and finding the minimum point.

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The 3 outside angles of a triangle need to equal 360.

n = 360 - 133 - 142 = 85

The answer is D.

divide the two numbers.

50/10 = 5

there are 5 campers per counselor, or

5/1 (camper/counselor)

hope this helps

50/10 or a simplified version 5/1 !! This Means For every 5 studens their are 1 counselor for them!

The width of the city is 104/5

Length of city is 5/2 times of width i.e. 5/2 × 104/5 = 520/10 = 52 metres

boat= 35mph

current=10mph

b = - [tex]\frac{1}{2}[/tex]

multiply both sides by 2(b + 1) to eliminate the fraction

3 = 3× 2(b + 1)

3 = 6(b + 1) = 6b + 6  ( subtract 6 from both sides )

- 3 = 6b ( divide both sides by 6 )

b = - [tex]\frac{3}{6}[/tex] = - [tex]\frac{1}{2}[/tex]

4 2 inch cubes are needed to fill the cubic box

The volume of the larger cube is 64 cubic inches and the volume of the smaller cube is 8 cubic inches. By dividing the larger volume by the smaller, we find that 8 two-inch cubes can fit in a four-inch cube.

The question is asking how many 2 inch cubes can fit into a bigger cube that has 4 inch edges. Firstly, we need to calculate the volume of the bigger cube by using the formula for volume of a cube which is the edge length cubed. So, 4 inches x 4 inches x 4 inches gives us a total of 64 cubic inches.

To find out how many 2 inch cubes can fit into the 64 cubic inch box, we would divide the volume of the bigger box by the volume of the smaller cube. The volume of the 2 inch cube is 2 inches x 2 inches x 2 inches which equals 8 cubic inches.

Dividing the volume of the bigger cube by the volume of the smaller cube gives us 64 divide by 8 equals 8. So, 8 cubes of 2 inch sides can fit into one cube of 4 inch sides.

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Answer:

5.6%

Step-by-step explanation:

The answer would be 1/8.

Answer:

it is 1905.88%

Step-by-step explanation:

We have,85%x X=1620

Or 85/100*x=1620

multiplying both sides by 100 and dividing both sides by 85

we have x = 1620 * 100/85

x=1905.88

if you are using a calculator ,simply enter 1620x100÷85,which will give you the answer

85 dinner rolls represent approximately 5.25% of 1620 dinner rolls. The calculation is done by dividing 85 by 1620 and then multiplying by 100.

To find what percent of 1620 dinner rolls is 85 dinner rolls, we use the basic formula of (part/whole) x 100. In this case, the 'part' is the 85 rolls, and the 'whole' is the 1620 rolls. So we calculate the percentage as follows:

(85 / 1620) x 100 = (0.05247) x 100 [10pt] [10pt]= 5.247%

Thus, 85 dinner rolls are approximately 5.25% of 1620 dinner rolls.

y - 2 = - 2(x - 7)

the equation of a line in ' point- slope form ' is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

to calculate m use the gradient formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (6, 4 ) and (x₂, y₂ ) = (7, 2)

m = [tex]\frac{2-4}{7-6}[/tex] = -2

using m = - 2 and (a, b) = (7, 2)

y - 2 = - 2(x - 7 ) ← in point-slope form

If your question is, "how much had Mr. Kork paid for the car?"

(8,400-200)/.4=$20,500

Achristensen's answer is correct. Just want to add a little more explanation how to arrive at the result.

let x be the original price.

Rewrite the verbal question as an equation using x:

8400 = 200 + 2/5 * x

solve for x:

8400-200 = 2/5*x

5/2*8200 = x

x = 20500

The original price was $20,500

the answer to this question is 21 dollars are left

Answer:

a) circumference of great circle=40212.4km

b) circumference of latutude circle=40212.4cos[tex]\theta[/tex] kilometers

c)The latitude circle is at an angle of 84.9°

d)Circumference of equator=40212.4km

Step-by-step explanation:

Refer to the figure shown below, we are given that earth has been charted with vertical and horizontal line and horizontal lines are called latitude lines.

So every horizontal line corresponds to a circle.

So the great circle will have radius equal to earth radius.

a)we are given with earth radius=6400km.

Hence circumference of a great circle= 2πX6400=40212.4km

b)At a latitude with angle x=[tex]\theta[/tex], the circumference of latitude circle = [tex]40212.4cos\theta[/tex]

c)If a latitude circle has a circumference of about 3593km,

then [tex]40212.4cos\theta=3593[/tex]

        [tex]cos\theta=\frac{3593}{40212.4} =0.0894[/tex]

         [tex]\theta=cos^{-1}(0.0894)[/tex]=84.9°

Hence the latitude circle is at an angle of 84.9°

d) At equator angle is 0.

Hence circumference of equator=40212.4cos(0)=40212.4km

Integer generally means number.

if s = seconds then

2s = 18

solve for s,

18/2 = 7

s = 7 seconds

Answer:

ball took 9 seconds to reach at the end o the ramp.

Step-by-step explanation:

Whitney rolls a ball down a ramp.

Length of the ramp = 18 feet

Speed by which the ball rolls down = 2 feet per second

Let the ball takes T seconds to reach the end of the ramp.

By the formula,

Time taken = [tex]\frac{\text{Distance rolled}}{\text{Speed}}[/tex]

                   = [tex]\frac{18}{2}[/tex]

                   = 9 seconds

Therefore, ball took 9 seconds to reach at the end o the ramp.

In 2009, there were 3 million more flex fuel vehicles than in 2004.

There is asked how many more flex fuel vehicles were available in 2009 compared to 2004. A flex fuel vehicle is one that can run on more than one type of fuel, usually gasoline combined with either ethanol or methanol fuel.

In 2004, there were approximately 5 million flex fuel vehicles. By 2009, this number rose to about 8 million. To find the increase, we subtract the initial number from the final number:

Number of flex fuel vehicles in 2009 - Number of flex fuel vehicles in 2004 = Increase in flex fuel vehicles

8 million - 5 million = 3 million