Can someone please help me with this table question . I don't understand :(

Mathematics

1 answer:

vichka [17]2 years ago

7 0

32. given (0,3)
fond slope by formula y2-y1/x2-x1 with any two points in the table ( you are lucky to be able to use formula in this equation!!) 5.3-3/1-0
2.3/1= 2.3 is slope you are able to check if this is true by doing it with every (x,y) point they give you.. luckyly the slope holds true for all the equation is
y=2.3x+3

33. this one the way I did it was plugging in (x,y) onto each equation and see of any hold true
y=4x+2.1
y=4x^2+2.1
y=2.1×4^x
y=4x

Let's plug in (1,8.4)
8.4=4 (1)+2.1
8.4=4+2.1
8.4=6.1 not true

8.4=4 (1)^2+2.1
8.4=4+2.1
8.4=6.1 not true

8.4=2.1×4^1 (1)
8.4=2.1×4
8.4=8.4 TRUE

8.4=4 (1)
8.4=4 not true

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How do you solve all of these ?

Flura [38]

5)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 5}}\quad ,&{{ -3}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies -1
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-3)=-1(x-5)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y+3=-x+5\implies y=-x+2$

6)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 3
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-1=3(x-2)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-1=3x-6\implies y=3x-5$

7)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 5}}\quad ,&{{ 2}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 0
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=0(x-5)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-2=0\implies y=2$

8)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -2}}\quad ,&{{ 0}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-0=\cfrac{5}{2}(x-(-2))
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=\cfrac{5}{2}(x+2)\implies y=\cfrac{5}{2}x+5$

9)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -2}}\quad ,&{{ 0}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-0=\cfrac{5}{2}(x-(-2))
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=\cfrac{5}{2}(x+2)\implies y=\cfrac{5}{2}x+5$

10)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{-2}}\quad ,&{{ -2}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-2)=\cfrac{3}{2}(x-(-2))
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y+2=\cfrac{3}{2}(x+2)\implies y+2=\cfrac{3}{2}x+3\implies y=\cfrac{3}{2}x+1$

4 0

2 years ago

he table gives the probability distribution of the number of books sold in a day at a bookstore. What is the probability of 16 o

Pepsi [2]

16 or more books sold in a day will include two intervals from the given table:

a) 16 - 20 books sold.
Probability = 0.201

b) 21 - 25 books sold.
Probability = ?

Since the table gives the probability distribution, the sum of all the probabilities must be equal to 1. We are given the probabilities of 4 out 5 intervals. So in order to find the probability of the 5th interval (21-25) we can subtract the sum of probabilities of rest of the intervals from 1.
So, P(21 - 25) = 0.019

In order to find the probability of 16 or more books sold we need to sum the probabilities of both these intervals.

So, the probability of 16 or more books being sold on a given day = 0.201 + 0.019 = 0.220

4 0

2 years ago

Please help me please

IgorLugansk [536]

The answer is 1, the solution is in the photo