All categories

3 years ago

How do you convert 1.66 into a fraction

Mathematics

2 answers:

joja [24]3 years ago

7 0

<span>=83/50 (as a improper fraction)
Proper fraction = 1</span> $\frac{33}{50}$

Svetlanka [38]3 years ago

4 0

1.66

=1.00+0.66

=100/100+66/100

=166/100

=83/50

You might be interested in

A chemist examines 15 geological samples for potassium chloride concentration. The mean potassium chloride concentration for the

NikAS [45]

Answer:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

Step-by-step explanation:

Data provided

$\bar X=0.376$ represent the sample mean for the chloride concentration

$\mu$ population mean (variable of interest)

s=0.0012 represent the sample standard deviation

n=15 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since we need to find the critical value first we need to begin finding the degreed of freedom

$df=n-1=15-1=14$

The Confidence is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value for this case with a t distribution with 14 degrees of freedom is $t_{\alpha/2}=2.145$

And the confidence interval is:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

7 0

3 years ago

Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0

Arada [10]

Answer:

$b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$

Step-by-step explanation:

The given matrix addition is

$B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

We need to find the elements of matrix B.

Let $B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}$

Substitute the value of matrix.

$\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$