Answer:
[tex]\filldraw (10,0) rectangle ++(3pt,3pt);[/tex]
Calculate the mean and median of: 202, 199, 223, 198, 223, 223, 301, 199.
Answer:
mean =225.25, median= 212.5
Step-by-step explanation:
You add all the numbers and divided by the total numbers to get mean, and for get the median you need to put them in order and get the number that is between all the number it be the 202 and 223 so you divide the two number with two and it will be 212.5
Pls give brainliest ily<3
Hello people ~
what's the sum of 7th odd number & 13th even number?
Answer:
13+26=39
Step-by-step explanation:
the 7th odd number is 13 and 13th even is 26
Answer:
[tex]\displaystyle 39[/tex]
Step-by-step explanation:
[tex]\displaystyle 39 = 13 + 26 \Rightarrow 39 = 13 + 2[13][/tex]
On a hundred chart, you will see that the seventh odd number is [tex]\displaystyle 13,[/tex]and the thirteenth even number is self-explanatory, so you should know what occurs there.
I am joyous to assist you at any time.
Phoebe spent $12.87 at the bookstore. She paid with a $20 bill. How much is her change?
Answer:
7.28 in change
Step-by-step explanation:
When she used 20 bucks it gave her 7.28 cents.
the rectangle shown has a perimeter of 56 cm and the given area. it's length is 8 more than three times it's width. write and solve a system of equations to find the dimensions of the rectangle.
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Let's solve ~
Assume width of rectangle be " x ", length = 3×width + 8 = 3x + 8 ~
Now, Perimeter of rectangle is :
[tex]\qquad \sf \dashrightarrow \:2(l + w) = 56[/tex]
[tex]\qquad \sf \dashrightarrow \:2(3x + 8 + x) = 56[/tex]
[tex]\qquad \sf \dashrightarrow \:2(4x + 8) = 56[/tex]
[tex]\qquad \sf \dashrightarrow \:4x + 8 = 56 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:4x + 8 = 28[/tex]
[tex]\qquad \sf \dashrightarrow \:4x = 28 - 8[/tex]
[tex]\qquad \sf \dashrightarrow \:x = 20\div 4[/tex]
[tex]\qquad \sf \dashrightarrow \:x =5 \: cm[/tex]
Hence, width = x = 5 cm
[tex]\qquad \sf \dashrightarrow \:l = 3w + 8[/tex]
[tex]\qquad \sf \dashrightarrow \:l = 3(5)+ 8[/tex]
[tex]\qquad \sf \dashrightarrow \:l = 15+ 8[/tex]
[tex]\qquad \sf \dashrightarrow \:l =23 \:cm[/tex]
And, length = 26 cm
») b. The area of the parallelogram is 65 ft2. What is its height? mm th 13 ft 13
Answer:
Five feet
Step-by-step explanation:
H = A ÷ L
H = 65 ÷ 13
H = 5 ft
Please I need answer quickly
how many ways can you form 3 groups from 18 people so that there's 3 people in first group, 6 people in the second group and 9 people in third
Answer:
4 084 080 ways
Step-by-step explanation:
The number of ways is :
[tex]C^{3}_{18}\times C^{6}_{15}\times C^{9}_{9}[/tex]
___________
Since
18C3=816
15C6=5 005
9C9=1
then
[tex]C^{3}_{18}\times C^{6}_{15}\times C^{9}_{9}=816\times5005\times1= 4\ 084\ 080[/tex]
The first term in a sequence is 84. The pattern follows the rule add 24. What are the next four terms?
Answer: 108, 132, 156, and 180
Step-by-step explanation:
If you need to add 24 to the first term for the next 4 terms, you would have
84 + 24 = 108
108 + 24 = 132
132 + 24 = 156
156 + 24 = 180
So your sequence would be 108, 132, 156, and 180 for the next 4 terms.
True or false the incenter of a triangle is the center of the only circle that can be circumscribed about it?
Answer:
true because the center of the triangle is in the middle
100 POINTS !! The trajectory of a potato launched from a potato cannon on the ground at an angle of 45 degrees with an initial speed of 70 meters per second can be modeled by the parabola f(x) = x − 0.002x2, where the x-axis is the ground. Find the height of the highest point of the trajectory and the horizontal distance the potato travels before hitting the ground. A.height: 125 m; distance: 500 m B.height: 250 m; distance: 22 m C. height: 500 m; distance: 125 m D. height: 22 m; distance: 250 m
Answer:
height: 125m; distance: 500m
Step-by-step explanation:
x = 250
f(x) = x − 0.002x2
f(x) = 250 − 0.002(250 )2
= 250 − 0.002(62,500)
= 250 − 125
= 125
250 times 2 = 500
[tex]\\ \rm\rightarrowtail y=x-0.002x^2[/tex]
So
Put 250[tex]\\ \rm\rightarrowtail y=250-0.002(250)^2[/tex]
[tex]\\ \rm\rightarrowtail y=250-0.002(62500)[/tex]
[tex]\\ \rm\rightarrowtail y=250-125[/tex]
[tex]\\ \rm\rightarrowtail y=125m[/tex]
This is max heightDistance:-
2×height2×125500mIn a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize
Answer:
Divide 10 by 25 then multiply by 4 to get 40 / 100 which gives you the percentage: 40%
11:16 1
(
61
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What is the area of this figure?
15 mi
4 mi
5 mi
11 mi
6 mi
4 mi
15 mi
18 mi
Write your answer using decimals, if
necessary.
Answer:
268
Step-by-step explanation:
Divide the figure into 3 pieces, a triangle, a rectangle, and another rectangle.
Triangle area=18*(15+6+1) * 1/2, or 189
Rectangle 1= 4*11= 44
Rectangle 2= 7*5= 35
Add all up, get 268
If the circumference of a circle is 220 cm, find its diameter and area
please correct answer
Step-by-step explanation:
C = 2 Pi R = 220
R = 220 / 2 pi = 35 cm
diameter = 35 x 2 = 70 cm
Area = pi r² = pi 35² = 3848.45 cm
maximum numbers of circles can be drawn through three noncollinear point is
Answer:
find the product of (4m-n)and(3m-2n)
Answer:
1
Step-by-step explanation:
the number of circles which can pass through three given non-collinear points is exactly one.
the reason is that the center of such a circle must be on the (perpendicular) bisectors of the lines between each pair of these points.
these bisectors intersect at one unique point which is the center of the circle and the distance of any point from the center is the radius. So given three non collinear points fixes the center and the radius thereby giving us one unique circle.
the same way we find the circumscribing circle of a triangle, which is exactly the same situation.
he negative square root of 16 is -4. What is the square root of -16?
About the minus sign with 16, square root of -1 is an imaginary number i(iota). Also we know that square root of -16 will be multiplication of square root of -1 and square root of 16. So the final answer becomes, + or - 4i. Hope this helps
the 6th term is 22 and the common difference is 6
Answer:
I have no idea what are you talking about......
which is the answers
Answer:
1. any number from 42 to 68
e.g 52
2. any number less than 42 or greater than 68
e.g 76
The volume of a right triangular prism is 72 cubic feet. the height of the prism is 9 feet. the triangular base is an isosceles right triangle. a right triangular prism has a volume of 72 cubic feet. triangles a e c and b f d are the base triangles. sides f d and b f are congruent and form a right angle. the height of the prism is 9 feet. what is the area of the triangular base? square feet what is the length of edge df? feet
Answer:
8 square feet
Step-by-step explanation:
The formula of a volume of a prism:
We have
Substitute:
divide both sides by 9
Answer:
8 and 4Step-by-step explanation:
The first one is 8
The second one is 4
MN is a diameter of a circle with centre ' O ' . If BD = CD , prove that ∠OAD = ∠OCD
~Please help!! Thanks in advance:)
Answer:
[tex] \sf \: Proved \: \angle \: OAD \: = \angle \: OCD[/tex]
Step-by-step explanation:
Given:
Mn is diameter of circle having centre O
and BD = OD,
To prove that:
[tex]\angle \: OAD \: = \angle \: OCD[/tex]
Solution:
Join the points O and B and draw OB,
On joining the line,
in ∆OCD and ∆OBD,
OC =OB → (Radius of same circle)
BD =CD → (from given)
OD =OD → (Common side in both the triangles)
Hence ∆OCD and ∆OBD are congruent from SSS property.
so we can say that,
[tex]\angle \: OBD \: = \angle \: OCD[/tex]
Consider above prove as statement A
Corresponding angles of congruent traingle.
in ∆ OAB,
OA = OB (radius of same circle)
hence ∆OAB is an isosceles traingle.
We know that opposite angle of isosceles traingle are always equal. hence,
[tex]\angle \: OBD \: = \angle \: OAB \\ \angle \: OAB \: = \angle \: OAD (same \: angles) \\ \angle \: OBD \: = \angle \: OAD[/tex]
Consider above prove as statement B
From Statement A & B we can say that
[tex]\angle \: OAD \: = \angle \: OCD[/tex]
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Find the missing dimension. Use the scale 1 : 15.
Actual Figure skater Height: 67. 5 in.
Answer:
4.5 inches
Step-by-step explanation:
1/15 of 67.5
This is if you are looking for the proportional height. That's all I can solve with given this info.
PLS HELP I am giving 25 points here is the question
A shopkeeper sold 12 bags of potatoes for $2.99 each. He bought the potatoes for $1.90 for each bag what was his total profit
Find theFind tha amount of profit per bag by subtracting:
2.99 - 1.90 = 1.09
He made $1.09 profit per bag.
Now multiply the profit per bag by the number of bags:
1.09 x 12 = 13.08
Total profit = $13.08 amount of profit per bag by subtracting:
2.99 - 1.90 = 1.09
He made $1.09 profit per bag.
Now multiply the profit per bag by the number of bags:
1.09 x 12 = 13.08
Total profit = $13.08
What is the longest line segment that can be drawn in a right rectangular prism that is 14 cm long, 11 cm wide, and 9 cm tall?
Answer:
the longest segment is between 14 and 9
Step-by-step explanation:
this is the best answer in my eyes
The length of the diagonal of the rectangular prism will be 19.95 centimeters.
What is the diagonal of the rectangular prism?A rectangle's longitudinal divides it into two equivalent right triangles. A rectangle's diagonal elements are all the same length. A rectangle is a rectangle whose diagonal elements bisect one another, cross at right angles, and are identical.
The length of the diagonal of the rectangular prism is given as,
D² = L² + W² + H²
The dimensions of the rectangular prism will be 14 cm long, 11 cm wide, and 9 cm tall.
The diagonal length is calculated as,
D² = 14² + 11² + 9²
D² = 196 + 121 + 81
D² = 398
D = 19.95 cm
The length of the diagonal of the rectangular prism will be 19.95 centimeters.
More about the diagonal of the rectangular prism link is given below.
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HELLPP
The rectangle shown is to be dilated by a scale factor of 3.5.
(a) Calculate the length of each side of the dilated image. Show your calculations for each side length.
(b) Draw the new image and label it
A B D C . (It does not have to be to scale)
Answer:
Answer:
long side/length is 63 cm and short side/height is 28 cm
Step-by-step explanation:
18 x 3.5 and 8 x 3.5
To dilate means to grow so you grow the side by 3.5
The base of a solid s is the region bounded by the circle x^2 y^2=16
The circle has a center at the origin (0, 0) and have a radius of 4 units.
What is a circle?A circle is the locus of a point such that its distance from a fixed point (center) is constant.
The standard equation of a circle is:
(x - h)² + (y - k)² = r²
Where (h, k) is the circle center and r is the radius.
The circle given by the equation:
x² + y² = 16
Hence the center is (0, 0) and the radius is 4 units.
The circle has a center at the origin (0, 0) and have a radius of 4 units.
Find out more on circle at: https://brainly.com/question/24375372
The center is (0, 0) and the radius is 4 units.
What is the circle?A circle is a round-shaped figure that has no corners or edges. In geometry, a circle can be defined as a closed, two-dimensional curved shape.
The base of a solid is s is the region bounded by the circle is;
[tex]\rm x^2+ y^2=16[/tex]
The standard equation of a circle is:
(x - h)² + (y - k)² = r²
Where (h, k) is the circle center and r is the radius.
On comparing with the standard equation of a circle;
h = 0 , k = 0 and r =4
Hence. the center is (0, 0) and the radius is 4 units.
Learn more on the circle at:
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Juwan and his friend Jamie are watching their favorite tech penny stocks in the stock market. The initial value of Jamie’s stock was $4.00, while Juwan’s stock’s initial value was $0.00. Because of some incredible turns in the market this morning, the stocks are now increasing in value. Every time Juwan’s stock rises by $4.00, Jamie’s stock increases by $3.00. If these stock gains remain constant throughout the day, perform the following:
a. Convert this scenario into two linear equations; show both the standard form and the slope-intercept form for both equations.
b. Explain how you decided to label the axes.
c. What are the realistic bounds for the domain and range of today’s stock gains? Explain your answer.
d. Does this scenario imply correlation or causation? Explain your answer.
Linear equations in slope intercept form for the given scenarios:
Juwan: y = 4x + 4
Jamie: y = 3x
Linear equations in standard form for the given scenarios:
Juwan: y - 4x = 4
Jamie: y - 3x = 0
Where, y = current stock value and x = number of times the stock value increased.
Domain: Set of natural number.
Range: Real number greater than or equal to 0.
This scenario is an example of causation.
What is standard form of line?The standard form of a line is Ax + By = C, where A and B are not both zero.
In the given scenario, we take
y = the present stock value of Jamie and Juwan.
x = number of times the stock value increased.
where, y depends upon x.
Juwan: y - 4x = 4
Jamie: y - 3x = 0
What is slope intercept form?The slope intercept form of a straight line is one of the most common forms used to represent the equation of a line. The slope intercept formula can be used to find the equation of a line when given the slope of the straight line and the y-intercept( the y-coordinate of the point where the line intersects the y-axis).
Juwan: y = 4x + 4
Slope = 4
y - intercept = 4
Jamie: y = 3x
Slope = 3
y - intercept = 0
What is domain and range?The domain is the set of all the input values of a function and range is the possible output given by the function.
In this particular question, x is the number of times stock price increased. Thus, it can be 0, 1, 2, 3....set of natural number. y is the current stock price. The stock price was positive in the beginning and has been increasing since then, therefore y lies in the set of positive real number.
What is causation?Causation means that one variable causes another to change, which means one variable is dependent on the other. Here, the price of stock depends upon the number of times its value increased.
Learn more about causation here
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solve 3x-1=5x+7
Please help
Answer:
[tex]3x - 1 = 5x + 7 \\ \\ 5x - 3x = 7 + 1 \\ \\ 2x = 8 \\ \\ x = \frac{8}{2} \\ \\ x = 4[/tex]
# Jerry Don is here#
what three ratios are equivalent to 6/18
Answer: 3:9, 1:3, 12:36
Step-by-step explanation: just multiply or divide your numerator and denominator by the same number. For the first one, I divided both 6 and 18 by 3, for the second I divided both 6 and 18 by 6, and for the last one I multiplied both 6 and 18 by 2. Hope this helps.
A tunnel for an amusement park ride has the shape of a
regular hexagonal prism with dimensions shown. The prism
has a volume of 3,572.1 cubic meters. Can two 8-meter cars
connected by a 3-meter connector pass through the tunnel
at the same time? Explain.
The 3,572.1 m³ volume of the hexagon and the 19 m. length of the cars and 3-m connector, gives;
Yes, two cars connected by a 3 meter connector can pass through the tunnel at the same timeHow can the capacity of the tunnel be found?From a similar question, we have;
Side length of the hexagon = 8.1 m
Perpendicular distance from the center to a side of the hexagon = 7 m.
Therefore;
Cross sectional area of the hexagon, A is found as follows;
A = 6 × 0.5 × 7 × 8.1 = 170.1
Length of the tunnel, D = 3572.1 ÷ 170.1 = 21
D = 21 meters
Length of two cars and a connector, L = 8 + 8 + 3 = 19
The tunnel length, D = 21 m. is longer than the length of two cars and the connector, L = 19 m.
Therefore;
Two cars connected by a 3 meter connector can pass through the tunnel at the same time.Learn more about the volume of a prism here;
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[tex]f(x) = 4x + 3[/tex]
[tex]g(x) = 5x - 4[/tex]
[tex]fg(x) = 20x + p[/tex]
Find the value of
[tex]p[/tex]
Let's see
[tex]\\ \rm\rightarrowtail (fog)(x)=f(g(x))[/tex]
[tex]\\ \rm\rightarrowtail f(5x-4)=20x+p[/tex]
[tex]\\ \rm\rightarrowtail 4(5x-4)+3=20x+p[/tex]
[tex]\\ \rm\rightarrowtail 20x-16+3=20x+p[/tex]
[tex]\\ \rm\rightarrowtail -13=p[/tex]
Answer:
[tex]\sf p = -13[/tex]
Step-by-step explanation:
Given functions:
[tex]\sf f(x)=4x+3[/tex]
[tex]\sf g(x)=5x-4[/tex]
[tex]\sf fg(x)=20x+p[/tex]
[tex]\sf fg(x)=f[g(x)][/tex]
This means g(x) is substituted for x in f(x).
[tex]\sf \implies f[g(x)]=4[g(x)]+3[/tex]
[tex]\sf =4(5x-4)+3[/tex]
[tex]\sf= 20x-16+3[/tex]
[tex]\sf =20x-13[/tex]
[tex]\sf f[g(x)]=f[g(x)][/tex]
[tex]\sf \implies 20x+p=20x-13[/tex]
[tex]\sf \implies p=-13[/tex]
A researcher examined a random sample of 300 homes in a small city and found that 47 had solar panels installed on their roofs. Use the sample to construct a 99% confidence interval for the proportion of all homes in the city that have solar panels installed on their roofs. What is the sample proportion? **Round to two decimal places
The confidence interval for the sample proportion is [tex]CI = 0.16 \pm 0.0545[/tex]
How to calculate the confidence interval for proportion?The formula for calculating the confidence interval for proportion is expressed as:
[tex]p \pm z\sqrt{p(1-p)/n[/tex]
p is the proportion
z is the z score at 99% interval
n is the sample size
Given the following parameters
[tex]p=\frac{47}{300} = 0.16\\z=2.576\\n=300[/tex]
Substitute the values
[tex]CI = 0.16 \pm 2.576\sqrt{0.16(0.84)/300}\\CI=0.16\pm2.576(0.02116)\\CI = 0.16 \pm 0.0545[/tex]
Hence the confidence interval for the sample proportion is [tex]CI = 0.16 \pm 0.0545[/tex]
Learn more on confidence interval here: https://brainly.com/question/15712887
Directions: Calculate the area of a circle using 3.14x the radius
1) d = 4.4 mm.
Calculate the area of the circle.
2) d = 3.7 cm.
Calculate the area of the circle.
3) r= 8.3 cm.
Calculate the area of the circle.
4) d = 5.8 yd.
Calculate the area of the circle.
5) d = 1 yd.
Calculate the area of the circle
6) r = 8 ft.
Calculate the area of the circle
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
As we know ~
Area of the circle is :
[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
And radius (r) = diameter (d) ÷ 2
[ radius of the circle = half the measure of diameter ]
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Problem 1[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 4.4\div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 2.2 \: mm[/tex]
Now find the Area ~
[tex]\qquad \sf \dashrightarrow \: \pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{} [/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
problem 2[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 3.7 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm[/tex]
Bow, calculate the Area ~
[tex]\qquad \sf \dashrightarrow \: \pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{} [/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2} [/tex]
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Problem 3[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 68.89[/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
Problem 4[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 5.8 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd[/tex]
now, let's calculate area ~
[tex]\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 8.41 [/tex]
[tex] \qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
problem 5[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 1 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd[/tex]
Now, let's calculate area ~
[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 0.25[/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
problem 6[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 64[/tex]
[tex]\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2} [/tex]
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