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HACTEHA [7]
3 years ago
7

Would this answer be Small sample ?

Mathematics
1 answer:
geniusboy [140]3 years ago
3 0

Yes, I would say the answer is Small Sample.

In order to make an accurate study, you would want to have more than 3 samples. I would say at least 500-1000 depending on what the subject is about and where it takes place.

Let me know if this helps :) Have a nice day!

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What is the value of 45 x 104 =<br> _?
Mashcka [7]

Answer:

4,680

Step-by-step explanation:

  1. 45 × 104 = 4,680

I hope this helps!

4 0
3 years ago
Read 2 more answers
What is the ratio of these plz help​
tatyana61 [14]

Step-by-step explanation:

1. Ratio(White to black) = 12 / [52 - 12]

Ratio(White to black) = 12 / [40]

Ratio(White to black) = 3 / 10 = 3:10

2. Ratio(Blue triangle to all) = 5 / [5+80]

Ratio(Blue triangle to all) = 5 / [85]

Ratio(Blue triangle to all) = 1 / 17 = 1:17

3. Ratio(Circle to all) = 76 / [22+76]

Ratio(Blue triangle to all) = 76 / [98]

Ratio(Blue triangle to all) = 38 / 49 = 38:49

4. Ratio(Yellow triangle to blue) = 18 / 14

Ratio(Yellow triangle to blue) = 9 / 7 = 9:7

5. Ratio(Yellow triangle to blue) = 42 / 6

Ratio(Yellow triangle to blue) = 7 / 1 = 7:1

6.Ratio(Boys to all) = 63 / [27 + 63]

Ratio(Boys to all) = 63 / 90

Ratio(Boys to all) = 7 / 9 = 7:9

6 0
3 years ago
Se desea constrir una barda utilizando block . las dimensiones de la barda son 25 m de largo y 2 m de altura . si se requieren 1
Andrew [12]

Answer:

I don't understand the language

Step-by-step explanation:

am not form Paris

3 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
Ethan says there are a total of 700 hundreds in the number 7,000,000.
maxonik [38]
Hello there.

Is Ethan correct?

No, Ethan is not correct. There are exactly 7,000 hundreds in 7,000,000.
6 0
3 years ago
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