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3 years ago

How do you create a model for 0.33 divided by 3

Mathematics

2 answers:

Triss [41]3 years ago

4 0

Since your chart is 100 squares you would want to shade 11 squares. Because your decimal is 11 out of 100.

Vlada [557]3 years ago

4 0

Shade 11 squares because that is 0.33/3 on the squares

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A first-degree equation in three variables graphs as a. a line c. a sphere b. a plane d. a paraboloid

Eva8 [605]

Answer:

plane

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The standard deviation of quiz scores is 5.6. Describe the quiz scores

Lyrx [107]

We have been given a table of quiz scores whose standard deviation is 5.6. We are asked to find the test scores that are within one standard deviation of the mean.

First of all, we will find the mean of the test scores using average formula.

$\text{Average}=\frac{\text{Sum of data points}}{\text{Number of data points}}$

$\text{Average}=\frac{25+13+16+30+27+22+19+22+15+28+27+29}{12}$

$\text{Average}=\frac{273}{12}$

$\text{Average}=22.75$

The scores within one standard deviation would be mean plus one SD and mean minus one SD that is:

$22.75-5.6\text{ and } 22.75+5.6$

$17.15\text{ and } 28.35$

Therefore, the quiz scores 17.15 and 28.35 are within one standard deviation of the mean.

7 0

4 years ago

Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5

givi [52]

<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

$A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]$ $X=\left[\begin{array}{c}x\\y\end{array}\right]$

and $B= \left[\begin{array}{c}11\\5\end{array}\right]$

∴AX=B

$adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]$

$A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6$

∴ $A^{-1} =\frac{adj A}{|A|}$

So, $A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}$

$A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}$

$X =A^{-1}\times B$

⇒ $\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]$