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Romashka-Z-Leto [24]
3 years ago
12

Eileen randomly chooses 15 seeds from the bag of flower seeds she has left over from last year, and 10 seeds from the bag she bo

ught this year. Which sampling technique did she use to collect the 25 seeds in her sample?
A.simple random sampling
B.systematic sampling
C.convenience sampling
D.stratified random sampling
Mathematics
2 answers:
amm18123 years ago
5 0
I believe it is D, stratified random sampling.
Hope this helps!!

harina [27]3 years ago
5 0
The correct answer would be D.).
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A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is
Illusion [34]

Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume V of a solid pyramid can be given as:

\displaystyle V = \frac{1}{3} \cdot b \cdot h,

where

  • b is the area of the base of the pyramid, and
  • h is the height of the pyramid.

Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be x cm.

  • At the vertice of this pyramid, x = 0 and the length of a side of the square is also 0.
  • At the base of this pyramid, x = 3 and the length of a side of the square is 4 cm.

As a result, the length of a side of the square will be

\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x.

The area of the square will be

\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}.

Integrate the area of the horizontal cross-section with respect to x

  • from the top of the pyramid, where x = 0,
  • to the base, where x = 3.

\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}.

In other words, the volume of this pyramid is 16 cubic centimeters.

5 0
3 years ago
Need help fast pls
DanielleElmas [232]
We're given the equation T = 3x + 2.
They're asking you to find the value of T when x = 1/3, so you all need to do is replace x by its value (in this case by 1/3) in the equation.

T = 3x + 2
T = 3 * (1/3) + 2
T = 3/3 + 2
T = 1 + 2
T = 3

So when x = 1/3 , T = 3.

Hope this Helps! :D
3 0
3 years ago
Read 2 more answers
A taxi service charges you an initial fee of $1.50 plus $1.50 for every mile traveled. A taxi ride costs $21.00. How many miles
Tasya [4]
the taxi driver when driver 5280 feet
6 0
3 years ago
Choose the correct trig ratio you would use to solve for the missing piece of the right triangle.
arlik [135]

Answer:

take 19 degree as reference angle

using sine rule

sin 19 degree=x/74

0.32=x/74

0.32*74=x

23.68=x

Step-by-step explanation:

Hope this helps u !!

8 0
2 years ago
Which two tables represent the same function?
topjm [15]

Answer:

The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

- There are five tables of functions, two of them are equal

- To find the two equal function lets find their equations

- The form of the equation of a line whose endpoints are (x1 , y1) and

  (x2 , y2) is \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

* Lets make the equation of each table

# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)

∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7

∴ \frac{y-8}{x-4}=\frac{7-8}{6-4}

∴ \frac{y-8}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 4) ⇒ simplify

∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

∴ x + 2y = 20 ⇒ (1)

# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ \frac{y-5}{x-4}=\frac{4-5}{6-4}

∴ \frac{y-5}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ \frac{y-8}{x-2}=\frac{5-8}{8-2}

∴ \frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

∴ x + 2y = 18 ⇒ (3)

# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

∴ \frac{y-10}{x-2}=\frac{14-10}{6-2}

∴ \frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1

- By using cross multiplication

∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

∴ -8 = x - y ⇒ switch the two sides

∴ x - y = -8 ⇒ (4)

# (x1 , y1) = (2 , 9) and (x2 , y2) = (8 , 6)

∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6

∴ \frac{y-9}{x-2}=\frac{6-9}{8-2}

∴ \frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 9) = -1(x - 2) ⇒ simplify

∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides

∴ x + 2y = 20 ⇒ (5)

- Equations (1) and (5) are the same

∴ The 1st and the 5th tables represent the same function

3 0
3 years ago
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