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lana [24]
4 years ago
11

PLEASE HELP ME!! The deepest point of Lake Titicaca in South America is -922 feet relative to its surface. The deepest point is

11,542 feet above sea level. What is the elevation of this lake?
Mathematics
1 answer:
miskamm [114]4 years ago
8 0

Answer:

12,464 ft

Step-by-step explanation:

I assume that the elevation of the lake is the altitude at the surface of the water.

The surface of the lake is 922 ft above the deepest point of the lake.

11,542 ft + 922 ft = 12,464 ft

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x1=-1.06956, x2=1.66215

Step-by-step explanation:

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(23.01) In a randomized comparative experiment on the effect of dietary calcium on blood pressure, researchers divided 58 health
Paul [167]

Answer:

113.7-2.154\frac{9.1}{\sqrt{29}}=110.06    

113.7+2.154\frac{9.1}{\sqrt{29}}=117.34    

And the 96% confidence is given by (110.06; 117.34)

Step-by-step explanation:

Information given

\bar X=113.7 represent the sample mean

\mu population mean (variable of interest)

s=9.1 represent the sample standard deviation

n=29 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)

The degrees of freedom are given by:

df=n-1=29-1=28

The Confidence is 0.96 or 96%, the significance is \alpha=0.04 and \alpha/2 =0.02, and the critical value would be t_{\alpha/2}=2.154

Replacing the info we got:

113.7-2.154\frac{9.1}{\sqrt{29}}=110.06    

113.7+2.154\frac{9.1}{\sqrt{29}}=117.34    

And the 96% confidence is given by (110.06; 117.34)

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3 years ago
The polygon is enclosed by a circle, center o so that each vertex touches the circumference of the circle
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1 year ago
T(v) = Av represents the linear transformation T. Find a basis for the kernel of T and the range of T.
lawyer [7]

Answer:

Ker(T) has basis \emptyset

Range(T) has basis \left\{\left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \right\}.

Step-by-step explanation:

If the matrix you wanted to represent  is A=\left(\begin{array}{cc}2&1\\3&4\end{array}\right), then:

ker(T)=\left\{{\bf x}\in \mathbb{R}^{2}: A{\bf x}=0\right\}.

So, to find ker(T) you must solve the homogenous equation

A{\bf x}=\left(\begin{array}{cc}2&1\\3&4\end{array}\right) \left(\begin{array}{c} x&y \end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)

Using Gauss elimination we obtain the simpler equivalent system

\left(\begin{array}{cc} -6&3\\0&5\end{array}\right) \left(\begin{array}{c} x&y\end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)

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x=0,y=0.

We have that ker(T)=\{\left(\begin{array}{c}0&0\end{array}\right)\}. On this case we say that the basis is the empty set \emptyset.

The range of T is the set of vectors of the form

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So,

Range(T)=\langle \left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \rangle. Where the angle brakets denotes the span.

5 0
3 years ago
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