Keegan fills a bucket of water, carries it to the flower bed and slowly pours the water over the new planted flowers. Select the graph that represents the height of the water in the bucket.
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Answer:
The first graph shows the height of water in the bucket
Related Questions
How do i solve and what’s the answer for number 14?
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Hello!
We have the following expression:
[tex]\sum_{n\mathop{=}0}^6300(1.05)^{n-1}[/tex]We can solve it as a geometric sequence, I'll show you how.
First, let's write the number of terms of the sequence:
Starting in 0 until 6, we have: a1, a2, a3, a4, a5, a6 and a7.
(7 terms).
Also, we can see that the ratio is 1.05.
Now, let's calculate the first term of the sequence (a1):
[tex]\begin{gathered} a_1=300\times(1.05)^{0-1} \\ a_1=300\times(1.05)^{-1} \\ a_1=300\times\frac{1}{1.05} \\ a_1=\frac{2000}{7} \end{gathered}[/tex]As we know the first term of the sequence and the ratio, we can use the formula below to calculate the sum of the 7 terms of this sequence:
[tex]\boxed{S_n=\frac{a_{1}(r^{n}-1)}{r-1}}[/tex]So, let's replace it with the values that we already know:
[tex]S_7=\frac{\frac{2000}{7}(1.05^7-1)}{1.05-1}=\frac{\frac{2000}{7}(1.4071-1)}{0.05}=\frac{\frac{2000}{7}(0.4071)}{0.05}\cong2326.29[/tex]Right answer: alternative B.need help asappppppp
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The pythagorean theorem is expressed as
hypotenuse^2 = opposite side^2 + adjacent side^2
Right triangles(triangles in which one of the angles is 90 degrees) have hypotenuse. Thus, the pythagorean theorem can only be used right triangles
Use the equation: …….d. Give the location of any oblique asymptote(s). If there is none, write “n/a.” (I only need help with option d)
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given
h(x) = x + 7/x^2 - 47
to locate if there is any oblque asymptotes
lim (x + 7/x^2 - 49/x)
There is no oblique asymptotes because the Dgree of numerator is less than that of denomator
Therefore the answer is n/a
Solve m+2n²=7m for n.
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We have to solve the n
1. subtract m of both sides of the expression:
[tex]m+2n^2-m=7m-m[/tex][tex]2n^2=6m[/tex]2. Divide both sides in 2
[tex]\frac{2n^2}{2}=\frac{6m}{2}[/tex][tex]n^2=3m[/tex]3. calculate the square root of both sides:
[tex]\sqrt{n^2}=\sqrt{3m}[/tex]Then:[tex]n^{}=\sqrt{3m}[/tex]John painted his most famous work, in his country, in 1930 on composition board with perimeter 107.69 in. If the rectangular painting is 5.14 in. Taller than it is wide, find the dimensions of the painting.
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Let w represent the width of the rectangular painting.
If the rectangular painting is 5.14 in. taller than it is wide, it means that
length = w + 5.14
The formula for calculating the perimeter of a rectangular is
Perimeter = 2(length + width)
Given that perimeter = 107.69, then
107.69 = 2(w + 5.14 + w)
107.69 = 2(2w + 5.14)
107.69 = 4w + 10.28
4w = 107.69 - 10.28 = 97.41
w = 97.41/4
w = 24.35
length = 24.35 + 5.14
length = 29.49
The dimesions are
width = 24.35 in
height = 29.49 in
Larry can spend at most $2800 to renovate his home. One roll of wallpaper costs $35, and one can of paint costs $40. He needs at least 20 rolls of wallpaper and at least 30 cans of paint. Identify the graph that shows all possible combinations of wallpaper and paint that he can buy. Also, identify two possible combinations.Possible combinations: (10, 10), (20, 15)Possible combinations: (25, 35), (40, 30)Possible combinations: (50, 40), (10, 60)Possible combinations: (11, 25), (18, 15)
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1) The best way to tackle this question is to set the inequalities then plot, and then we can check.
2) Gathering the data we can write out the following inequalities:
[tex]\begin{gathered} 35w+40c\leq2800 \\ w\ge20 \\ c\ge30 \end{gathered}[/tex]Notice that "w" stands for rolls of wallpaper and "c" cans of paint.
3) Let' plot them and in the darker region we'll see where are the possible solutions, i.e. combinations:
Note that this yellow part is the area where are located those possible solutions.
Let's plot the options to check them within the graph:
Note that among the options the only one that is located in the darkest region is the (25, 35)
can you please help me with this
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Option B
Explanations:Note that:
perpendicular lines are lines that intersect at 90 degrees
That is, two lines that are perpendicular will form a right angle
The only figure that contain perpendicular lines is the figure in option B. Each of the lines that make up the figure intersect at right angle
chris and sue went to the Chinese restaurant down the street. Their bill was $25.00 and they gave 22% tip. After they calculated the total, they split the bill evenly. How much did each person pay?
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The bill was $25.00 and they gave a 22% tip.
First, let us find 22% of $25.00
[tex]tip=\frac{22}{100}\times25.00=5.50[/tex]The total bill is
[tex]total\: bill=\$25.00+\$5.50=\$30.50[/tex]They split the bill evenly so divide the total bill by 2.
[tex]\frac{\$30.50}{2}=\$15.25[/tex]Therefore, each person paid $15.25.
Chris = $15.25
Sue = $15.25
Solve for xX - 4 =(2x - 5Step 1 of 5Using the squaring property of equality, eliminate the radical from the equation.Square both sides of the equation to eliminate the square root. Then, expand the square of the binomial usingthe formula, (x - y)2 = x2 - 2xy + y?(x - 4)2 =(v2x - 5)(x - 4)2 = 2x -x² - 2(x)1) + 42 = 2x -X + 16 = 2x -
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Step 1
Expand both sides of the equation.
[tex]\begin{gathered} (x-4)^2=(\sqrt[]{2x-5)}^2 \\ (x-4)\text{ ( x-4) = (}\sqrt[]{2x-5)}\text{ (}\sqrt[]{2x-5)} \end{gathered}[/tex][tex]\begin{gathered} x^2-4x-4x\text{ +16 = }(2x-5)^{\frac{1}{2}}(2x-5)^{\frac{1}{2}} \\ x^2-8x+16=(2x-5)^{\frac{1}{2}+\frac{1}{2}} \end{gathered}[/tex][tex]\begin{gathered} x^2-8x+16=(2x-5)^1 \\ x^2-8x\text{ +16 = 2x-5} \end{gathered}[/tex]Step 2
Apply the given formula to the answer in step 1
[tex]\begin{gathered} y\text{ }=\text{ 4} \\ x^2_{}-2(x)(4)+4^2=2x-5 \end{gathered}[/tex]Hence the required answer is
[tex]x^2-2(x)(4)+4^2=2x-5[/tex]if f(x) =(4x^2 -11)^3 and g(x) = 4x^2 -11. given that f(x) = (h°g)(x) find h(x)
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Given:
[tex]\left(x\right)=\left(4x^2-11\right)^3andg\left(x\right)=4x^2-11.[/tex]Required:
Find h(x) if
[tex]f\mleft(x\mright)=h°g\left(x\right)[/tex]Explanation:
[tex]\begin{gathered} f\mleft(x\mright)=h°g\left(x\right) \\ f\mleft(x\mright)=h(g(x)) \end{gathered}[/tex]Let g(x) = x
[tex]f(x)=h(x)[/tex][tex](4x^2-11)^3=x^3[/tex]Solve by taking cube root on both sides.
[tex]x=4x^2-11[/tex][tex]\begin{gathered} g(x)=x \\ g(x)=4x^2-11 \end{gathered}[/tex]A collection of points is shown on the complex plane.Which point represents z1z2?
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Given:
[tex]\begin{gathered} z_1 \\ z_2 \end{gathered}[/tex]You need to remember that a Complex Number has this form:
[tex]a+bi[/tex]Where "a" is the real part and this is the imaginary part:
[tex]bi[/tex]You can identify that:
[tex]z_2=-4+0i[/tex]And:
[tex]z_1\approx-2.8+2.8i[/tex]You need to multiply them in order to find:
[tex]z_1z_2[/tex]You get:
[tex]z_1z_2=(-4+0i)(-2.8+2.8i)[/tex][tex]z_1z_2=(-4)(-2.8+2.8i)[/tex][tex]z_1z_2=(-4)(-2.8)+(-4)(2.8i)[/tex][tex]z_1z_2=11.2-11.2i[/tex]You can identify that the approximate coordinates of point R are:
[tex]R(11.2,11.2i)[/tex]Hence, the answer is: Third option.
ScenarioMeagan will be using a moving truck to move her belongings to another location. She is trying todecide whether to rent Truck A or Truck B:Option 1: Truck A offers a flat fee of $35 and $0.45 per mile.• Option 2: Truck B offers a flat fee of $40 and $0.25 per mileAnswer the following questions:
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Given
Distance needs to drive the truck = 15 miles
Option 1: Truck A offers a flat fee of $35 and $0.45 per mile.
Option 2: Truck B offers a flat fee of $40 and $0.25 per mile.
Find
a) Cost of using truck A
b) Cost of using truck B
c) Which Company offers the best deal at this mileage
Explanation
Let number of miles driven = x
then the linear equation that gives the cost of renting the truck A is
Cost = $35 + $0.45(x)
linear equation that gives the cost of renting the truck B is
Cost = $40 + $0.25(x)
If the truck is driven 15 miles , that is x = 15 miles , then
cost of using truck A =
[tex]\begin{gathered} 35+0.45(15) \\ 41.75 \end{gathered}[/tex]and cost of using truck B =
[tex]\begin{gathered} 40+0.25(15) \\ 43.75 \end{gathered}[/tex]Final Answer
a) cost of using truck A = $ 41.75
b) cost of using truck B = $ 43.75
c) Truck A offers the best deal at this mileage
can you help me solve for Z 12z=108
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We have the equation:
[tex]\begin{gathered} 12\cdot z=108 \\ \text{Divided all by 12, we have:} \\ \frac{12\cdot z}{12}=\frac{108}{12} \\ z=\frac{108}{12}=9 \end{gathered}[/tex]To solve this equation we just need to isolate z, in this case we pass the 12 dividing.
Situation #5Mrs. Jimenez makes and sells monogrammed fleeceblankets. It costs her $5.95 for the fabric for each blanketand $175 per month to rent the embroidery machine. Shesells the blankets for $29.95 each. Write an equation orinequality to find b, the number of blankets Mrs. Jimenezmust sell to earn a profit each month.
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Given in the question that;
cost of fabric fr each blanket is = $5.95
cost of renting the machine is per month is =$175
selling price per blanket is $29.95
The number of blankets she must sell to earn a profit , b , will be represented by the equation or inequality as;
Cost of producing the blanket = $5.95 + $175 = $180.95
Cost of selling a blanket = $29.95
The number of blankets to sell so as to cover production costs will be;
$180.95/$29.95= 6.04
This is approximately 6 blankets
The minimum number of blankets to sell should be 6 blankets
So for a profit, the number of blankets must be more than 6
Hence ; b> 6
Answer is : b > 6
I) what is the ratio of their volume?II) write an expression in terms of r for the volume inside the sphere but outside the cone.
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We will have the following:
First, we can see that the volume of the sphere and the regular cone are given by:
[tex]V_s=\frac{4}{3}\pi r^3[/tex]And:
[tex]V_c=\frac{1}{3}\pi r^2\ast h[/tex]Now, since the volume of the cone is inscribed by a maximum stablished by the sphere, we know that the maximum height for the cone will be equal to the radius of the sphere, so, we re-write the volume of the cone:
[tex]V_c=\frac{1}{3}\pi r^2\ast r\Rightarrow V_c=\frac{1}{3}\pi r^3[/tex]i. Now, we determine the ratio of both volumes as follows:
[Cone to sphere]
[tex]\begin{gathered} r=\frac{(1/3)\pi r^3}{(4/3)\pi r^3}\Rightarrow r=\frac{(1/3)}{(4/3)} \\ \\ \Rightarrow r=\frac{1\ast3}{4\ast3}\Rightarrow r=\frac{1}{4} \end{gathered}[/tex]So, the ratio of the volumes of the cone to the sphere will be of 1:4.
ii. And the expression in terms of r for the volume inside the sphere but outside the cone will be:
[tex]V=\frac{4}{3}\pi r^3-\frac{1}{3}\pi r^3\Rightarrow V=\pi r^3[/tex]So, the expression is:
[tex]V=\pi r^3[/tex]7 pounds of potatoes cost $63. How many pounds ofpotatoes can you get with $261?
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We know that 7 pounds of potatoes cost $63.
This is a proportional relationship between quantity (pounds of potatoes) and cost.
We can calculate how many pounds of potatoes we can get with $261 applying the Rule of three:
[tex]261\text{ \$}\cdot(\frac{7\text{ pounds}}{63\text{ \$}})=29\text{ pounds}[/tex]We can buy 29 pounds of potatoes with $261
NOTE: I use the units ($ and pounds) to know how to multiply them in order to get the result I was looking for.
When it gets hot, children play in Burlington's circular water fountain. According to the architectural plans, it has a circumference of 119.32 meters. What is the fountain's radius?
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Circumference of a circle formula
[tex]C=2\pi r[/tex]where r is the radius of the circle.
Substituting with C = 119.32 meters and solving for r:
[tex]\begin{gathered} 119.32=2\pi r \\ \frac{119.32}{2\pi}=\frac{2\pi r}{2\pi} \\ 19\text{ meters }\approx r \end{gathered}[/tex]Richard took a group of friends and their kids to see a Star Wars movie in 3D. He bought a total of 20 tickets. Adult movie tickets were $12 and children's tickets were $8. If the total cost of the tickets was $192, how many children's tickets did Richard buy?
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Data:
Total tickets: 20
Adult movie tickets: A
Children's tickets: C
A: $12
C: $8
As the total number of tickets is 20 you have the next.
[tex]A+C=20[/tex]As the total cost of the tickets is $192 you have the next:
[tex]12A+8C=192[/tex]Then, you have the next system of equations:
[tex]\begin{gathered} A+C=20 \\ 12A+8C=192 \end{gathered}[/tex]To solve a system of equations:
1. Solve one of the variables in one of the equations:
Solve A in the first equation:
[tex]A=20-C[/tex]2. Use the value of A you get in the first part in the other equation:
[tex]12(20-C)+8C=192[/tex]3. Solve C:
- Remove parenthesis. Distributive property:
[tex]240-12C+8C=192[/tex]- Combine like terms:
[tex]240-4C=192[/tex]- Substract 240 in both sides of the eqaution:
[tex]\begin{gathered} 240-240-4C=192-240 \\ \\ -4C=-48 \end{gathered}[/tex]-Divide both sides of the equation into -4:
[tex]\begin{gathered} \frac{-4}{-4}C=\frac{-48}{-4} \\ \\ C=12 \end{gathered}[/tex]As C is 12. Richard bought 12 Children's tickets
Identify the missing symbol.0.8 ? 3/5 A. = B. > C.
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This exercise can be solved by substracting the number on the right from that on the left. Then we have three possibilities:
If the result of this substraction is 0 then the correct symbol is =.
If the result is greater than 0 then the correct symbol is >.
If the result is smaller than 0 then the correct symbol is <.
Then we get:
[tex]0.8-\frac{3}{5}=0.2[/tex]Answer0.2 is greater than 0 which means that the correct symbol is >. Therefore the answer is option B.
Which of the following statements are true?Select all that apply.A. 9 is a perfect cube.B. 27 is a perfect cube.C. 10 is neither a perfect square nor a perfect cube.D. 25 is a perfect square.OE. 2,744 is both a perfect square and a perfect cube.
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A perfect square is the square of a whole number:
D. 25 is a perfect square. True , because 5x5 =25
A perfect cube is the cube of a whole number:
B. 27 is a perfect cube. True, Because 3x3x3 = 27
The rest if False
Complete the measure of the interior angles of the given triangle.
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It says that the given triangle is an isosceles triangle, which means that two of its sides are congruent.
As shown in the figure, the sides across ∠A and ∠C are congruent. This condition must also mean that ∠A and ∠C are congruent.
[tex]\angle A\text{ = }\angle C[/tex]∠C and its adjacent angle measuring 130° are Supplementary, which means that their sum is equal to 180°.
With this, we can get the measure of ∠C.
[tex]\begin{gathered} \angle C+130^{\circ}=180^{\circ} \\ \angle C+130^{\circ}-130^{\circ}=180^{\circ}-130^{\circ} \\ \angle C=180^{\circ}-130^{\circ} \\ \angle C=50^{\circ} \end{gathered}[/tex]But,
[tex]\angle A\text{ = }\angle C[/tex]Therefore, ∠A must also be equal to 50°.
[tex]\angle A\text{ = }\angle C=50^{\circ}[/tex]The total sum of all interior angles of a triangle is 180°. With that relationship, we can determine the measure of ∠B since we've already determined the measure of ∠A and ∠C.
We get,
[tex]\angle A\text{ + }\angle B\text{ + }\angle C=180^{\circ}[/tex][tex]\begin{gathered} \text{ 50}^{\circ}\text{ + }\angle B+50^{\circ}=180^{\circ} \\ \angle B+100^{\circ}-100^{\circ}=180^{\circ}-100^{\circ} \end{gathered}[/tex][tex]\angle B=80^{\circ}[/tex]Therefore, ∠A = 50°, ∠B = 50° and ∠C = 80°.
The probability that an employee will be late to work at a large corporation is 0.21. What is the probability on a given day that in a department of 5 employees 1 or 2 employees are late ?
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Let the probability that an eployee will be late to work be given by:
[tex]P(1|1)=0.21[/tex]Then, in a department of 5 employes, the problability that 1 or 2 employes are late is given by:
[tex]\begin{gathered} P((1\lor2)|5)=P(1|5)+P(2|5) \\ P(1|5)=P(1|1)\cdot(1-P(1|1))^4\cdot\frac{5!}{1!\cdot4!}=0.21\cdot0.79^4\cdot5=0.41 \\ P(2|5)=P(1|1)^2\cdot)(1-P(1|1))^3\cdot\frac{5!}{2!\cdot3!}=0.21^2\cdot0.79^3\cdot10=0.22 \\ \therefore P((1\lor2)|5)=0.41+0.22=0.63 \end{gathered}[/tex]A dart player made 80 bull's eyes out of 100 attempted throws. What percent of the throws were bull's eyes?
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To find the percentage, we just have to divide.
[tex]\frac{80}{100}=0.80[/tex]Hence, 80% of the attempts were bullseyes.2,399.1 in scientific notation
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We are given the following number
[tex]2399.1[/tex]We want to write it in scientific notation
How to convert to scientific notation:
1. Move the decimal point to the left until a single digit is left (which means stop after 3)
2. Count the number of times you moved the decimal point (that's your exponent the power of 10)
3. Finally write it in the standard format
The number in the scientific notation is
[tex]2.3991\times10^3[/tex]We moved the decimal point to the left (3 times) until only a single digit is left.
The exponent is 3 (power of 10)
The exponent is positive when we move to the left.
The exponent is negative when we move to the right.
7-4) Find all possible values of a^3 + b^3 if a^2 + b^2 = ab = 4
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The given expression is:
[tex]a^3+b^3[/tex]This can be expressed as:
[tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]This can also be written as:
[tex]a^3+b^3=(a+b)(a^2b_{}+b^2-ab)[/tex]Since a² + b² = ab:
a² + b² - ab = 0
Therefore:
[tex]\begin{gathered} a^3+b^3=(a+b)(0) \\ a^3+b^3=0 \end{gathered}[/tex]The value of a³ + b³ if a² + b² = ab = 4 is 0
Find the value of z for the normal distribution such that 0.08 of the area lies to the right of z.
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The standardized normal distribution has the following shape:
The horizontal axis is the z-value. The graph opens in both left and right ends to minus infinity and infinity, respectively, and the area under the whole curve is 1. Areas under the graph represent the probability that z takes certain values. (between two specific values, higher tan some value, or lower than some value).
Normal distribution tables allow us to calculate the area under the curve at the left of certain values of z. We are asked to find a value of z such that the area at its right is 0.08:
Usually, tables give values of z at the left of some value, so we need to rewrite the question. The total area under the curve is 1, and the area at the right of z is 0.08. This means that the area at the left of z is:
[tex]1-0.08=0.92[/tex]Then, we can say that we need to find the value of z that has an area at is left of 0.92.
If we read from a table of z-values, we can find that the value with an area of 0.92 at its left is approximately 1.4.
The two shapes are similar. Triangle ZCW is the original what is the scale factor Z:(0,4) W:(-4,0) C:(4,-4) R:(0,3) J:(-1,2) Q:(1,1)
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Let's find the length of one of the sides of WZC and one of the sides of JRQ:
[tex]\begin{gathered} WZ=\sqrt[]{(-4-0)^2+(0-4)^2}=\sqrt[]{32}=4\sqrt[]{2} \\ JR=\sqrt[]{(0-(-1))^2+(3-2)^2}=\sqrt[]{2} \\ WZ\cdot k=JR \\ 4\sqrt[]{2}\cdot k=\sqrt[]{2} \\ k=\frac{\sqrt[]{2}}{4\sqrt[]{2}} \\ k=\frac{1}{4} \end{gathered}[/tex]Jeric started taking up his workout session in a gym. On the first week, he workout for 40 minutes per day, on the second week he workout for 50 minutes per day. Each week, he wants to increase his working out session by 10 minutes per day. If he workout for five days each week. What will be his total jogging time on seventh week? • 450 minutes• 400 minutes• 500 minutes• 900 minutes
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We can solve this as a series;
40,50,60,70,80,90,100.
On the second week, she worked out for 100 minutes per day.
Since he works out 5 days a week.
His total jogging time on the seventh week would be 5 x 100 = 500 minutes.
A tree casts a shadow 24 feet long. At the same time, a vertical rod 24 feet high casts a shadow 16 feet long. How tallis the tree?
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Let the height of the tree be 'h'
We can say:
h casts a shadow of 24
24 casts a shadow of 16
We can put it into a ratio:
[tex]\frac{h}{24}=\frac{24}{16}[/tex]Now, after cross multiplying, we can easily figure out the value of the variable "h", the height of the tree. Steps are shown:
[tex]\begin{gathered} \frac{h}{24}=\frac{24}{16} \\ 16h=24\times24 \\ 16h=576 \\ h=\frac{576}{16} \\ h=36 \end{gathered}[/tex]The tree is 36 feet tall.
Answer:
36 feet95-85-닚Time (seconds)a01120 140 160 180 200Weight (pounds)Nineteen people ran 400 meters. Each point on the scatterplot above represents a person's time (inseconds) to complete the run and the weight (in pounds) of that person. The scatterplot also shows a lineof best fit for the data. According to the graph, each increase of 8 pounds in a runner's weight tends toincrease the runner's time by approximately how many seconds?2.002.53.03.54.0
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The best fit line passes throug the points (120,65) and (200,90).
Determine the slope of best fit line.
[tex]\begin{gathered} m=\frac{90-65}{200-120} \\ =\frac{25}{80} \end{gathered}[/tex]So for unit increase in weight time increase by 25/80 se