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Alex777 [14]
3 years ago
10

Smart phone users between 18 and 24 years of age send on average 64 text messages per day with a standard deviation of 10 messag

es about what percentage of smartphone users in this age groups in less than 51 text messages per day
Mathematics
1 answer:
inna [77]3 years ago
4 0

Answer:

9.68%

Step-by-step explanation:

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Marisa and Sandy ran as far as they could in 30 minutes. Sandy ran 2.5 km farther than Marisa. They ran a total distance of 9.5
Ivan
Sandy ran 6 km, and Marisa ran 3.5 km.

6 + 3.5 = 9.5


I really hope this helped you out! c:
7 0
3 years ago
Read 2 more answers
Cw 11.1 backside v. V.
nika2105 [10]

Answers:

10.  28.3 cm²

11. 7.3 cm²

12. P: 32 yds, A: 76.8 sq yds

13. P: 64 units, A: 220.8 units

14.  50.3 cm²

15. 132.7 cm²

16. 9.7 in

17. 5.9 cm

18.  324.08 sq inches.

19. 39 cm²

20. 36.6 cm²

Step-by-step explanation:

10. Find the area of the shaded region.

It's a circle, let's first calculate its area, with the formula A = π * r²

A = π  * 6² = 36 π  = 113.1 cm²

The shaded area has a 90 degrees angle... so it's 1/4 of the whole circle.

The area of the shaded area of this circle is then: 113.1 * 1/4 = 28.275 cm²

Rounded to the tenths, that's of course 28.3 cm²

11. Find the area of the shaded region.

Again, let's first find the area of the complete circle, with A =  π * r²

A =  π * 4² = 16π  = 50.27 cm²

Now, let's find the angle for the shaded area.

A circle has 360 degrees, and we know the angles of 2 segments... so we can easily find the missing angle:

360 - 203 - 105 = 52

The shaded area has an arc of 52 degrees, so we multiply the area of the full circle by 52/360 to get the area of that shaded area:

(50.27 * 52) / 360 = 7.26 cm², rounded to the tenths: 7.3 cm²

12. Find perimeter AND area of this regular polygon.

The figure is a regular octagon (8 sides).

To calculate its perimeter, it's simply 8 times one side, so: 8 * 4 yd = 32 yds

For the area, you can view an octagon as 8 triangles joined together.  In this case, we have a base of 4 yds and a height of 4.8 yds, so the area of each triangle is: (4 * 4.8) /2 = 9.6, the total area of the octagon is then 8 * 9.6 = 76.8 yds

13. Find perimeter AND area of this regular polygon.

The figure is a regular octagon (8 sides).

To calculate its perimeter, it's simply 8 times one side, so: 8 * 8 = 64 units

For the area, you can view an octagon as 8 triangles joined together.  In this case, we have a base of 4 yds and a height of 4.8 yds, so the area of each triangle is: (8 * 6.9) /2 = 27.6, the total area of the octagon is then 8 * 27.6 = 220.8 units

14. Find the area of a circle of <u>radius</u> = 4 cm

We have a circle with a radius of 4 cm, we need to find its area.

The area of a circle is obtained by the formula: A = π * r²

We already have the value of r, so we will input it in the formula:

So, we'll have A = π * 4² = 16 * π = 50.27 cm²

Rounded to the tenths: 50.3 cm²

15. Find the area of a circle of <u>diameter</u> 13 cm

We have a circle with a diameter of 13 cm, we need to find its area.  To use the formula for the area, we need the radius, not the diameter.  Since the diameter is 13 cm, the radius is 6.5

The area of a circle is obtained by the formula: A = π * r²

So, we'll have A = π * 6.5² = 42.25 * π = 132.73 cm²

Rounded to the tenths: 132.7 cm²

16. Find the diameter of a circle with an area of 75 in².

We just used the area formula for a circle, based on the radius.  We'll process it in reverse from the area to get the radius... which will give us the diameter, so r² = A / π

We then have:  r² = 75 / π = 23.87

So, r = 4.88 (square root of 23.87), which we double to get the diameter: 9.76 in, rounded to 9.7 in.

17. Find the radius of a circle with an area of 108 cm².

We just used the area formula for a circle, based on the radius.  We'll process it in reverse from the area to get the radius, so r² = A / π

If we input the area given in the question into the formula, we have:  r² = 108 / π = 34.38 cm

So, r = 5.86 (square root of 34.38), rounded to 5.9 cm.

18. Find the area

We have here a half-circle with a triangle.

We just used the formula for the area of a circle, and we know the radius of that half-circle: 8 inches.

A = π * 8² = 64 * π = 201.06 sq inches, for the whole circle.

But we only have half of it, so 201.06 / 2 = 100.08 sq inches.

Now the triangle.  To calculate the area of a triangle, you multiply its base by its height, then you divide by 2, so:

A = 16 x 28 = 224 sq inches.

Which we add to the half-circle area (100.08 sq inches) to get 324.08 sq inches. So 324.1 sq inches once rounded to the tenths.

19. Find the area

We have a triangle on top of a rectangle.

Area of a triangle: (base * height) / 2.

Area of a rectangle: base * height

So, for the triangle: A = (6 * 3) / 2 = 9 cm²

And for the rectangle: A = 6 * 5 = 30 cm²

Total: 9 + 30 = 39 cm²

20. Find the area

We have a rectangle and a half circle.  The radius of the half-circle is: 2 cm (half of the height of the rectangle).

Area of the rectangle: base * height, so 6 * 4 = 24 cm²

Area of the circle:  π * r² = π * 2² = 4π = 12.56 cm²

Total for the figure: 24 + 12.56 = 36.56, or 36.6 cm² once rounded.

3 0
3 years ago
Once you have a single variable isolated in one of the two equations in a system, the second step of using substitution to solve
Ket [755]
I think C, not sure though :)
6 0
4 years ago
Read 2 more answers
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

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3 years ago
Write an inequality to represent the graph.<br> (0,2) (4,-1)
dusya [7]

Answer:

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Step-by-step explanation:

8 0
3 years ago
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