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zhenek [66]
3 years ago
6

Suppose that the mean of the sampling distribution for the difference in two sample means is 0.

Mathematics
1 answer:
Simora [160]3 years ago
8 0

Answer:

Optin b is right

Step-by-step explanation:

given that the  mean of the sampling distribution for the difference in two sample means is 0.

This means that when two sample means are equal we get difference is 0 and this would be the point estimate of mean for differene of means distribution.

Hence the four options given can be analysed

. the two sample means are both 0.   -- False need not be 0 but equal

B. the two sample means are equal to each other.   -- True

C. the two population means are both 0.  -- False need not be 0

D. the two population means are equal to each oth  -- False only sample means are 0 and if sample represents population this may be 0

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Write an absolute value equation for the graph shown below
BabaBlast [244]
So, hmmm keeping in mind that, an absolute value expression is in effect a piece-wise function, let's find its "slopes", the negarive slope on the left-side and the positive slope on the right side, and let's use the vertex point as our first point

what is the slope of a line that passes through (2, -1) and (0,7)

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ 2}}\quad ,&{{ -1}})\quad 
%   (c,d)
&({{ 0}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{7-(-1)}{0-2}\implies \cfrac{7+1}{-2}\implies -4
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=-4(x-2)
\\\\\\
\boxed{y+1=-4(x-2)}

now, what is the slope of a line that passes through (2, -1) and (4,7)

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ 2}}\quad ,&{{ -1}})\quad 
%   (c,d)
&({{ 4}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
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% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=4(x-2)
\\\\\\
\boxed{y+1=4(x-2)}

since an absolute value expression is just a piece-wise with a +/- versions, then we can just combine those two.

\bf \begin{cases}
y+1=-4(x-2)\\
y+1=4(x-2)
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All the answers are in the link

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Which expression is equivalent to 3(5 + 4)? <br> 15 + 12 <br> 15 + 4 <br> 8 + 7 <br> 8 + 4
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Find the absolute maximum and absolute minimum values of f on the given interval.
anyanavicka [17]

The question is missing parts. Here is the complete question.

Find the absolute maximum and absolute minimum values of f on the given interval.

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              Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

<u>Absolute</u> <u>maximum</u> is a point where the function has its greatest possible value.

<u>Absolute</u> <u>minimum</u> is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

f(x)=xe^{-\frac{x^{2}}{32} }

Using the product rule, first derivative will be:

f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )

f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} ) = 0

1-\frac{x^{2}}{16}=0

\frac{x^{2}}{16}=1

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x can't be -4 because it is not in the interval [-2,8].

f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76

f(4)=4e^{-\frac{4^{2}}{32} }=2.42

f(8)=8e^{-\frac{8^{2}}{32} }=1.08

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

8 0
3 years ago
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