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RSB [31]
3 years ago
9

#4 home do I find which one it is?

Mathematics
1 answer:
lina2011 [118]3 years ago
8 0
\bf \begin{array}{lllll}
solutions&graphs&slopes\\
----&----&----\\
\textit{exactly one}&
\begin{array}{llll}
\textit{the two lines intersect}\\
\textit{at one point}
\end{array}&\textit{different slopes}\\\\
infinitely\quad many&
\begin{array}{llll}
\textit{the two lines coincide}\\
\textit{one is right on top}\\
\textit{ of the other}
\end{array}&
\begin{array}{llll}
\textit{equal slopes}\\
\textit{equal y-intercepts}
\end{array}
\end{array}

\bf \textit{no solution}\qquad\quad &\textit{lines are parallel} \qquad &
\begin{array}{llll}
\textit{equal slopes}\\
\textit{different y-intercepts}
\end{array}
\end{array}

for example, let's look at the first set

y+3x =5   or  y = -3x+ 5
and               y = -3x + 2
                    y =  m  + b

the slopes are equal, the y-intercepts differ
that means, they're just parallel lines, no solution
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If y varies directly as x, and y is 18 when x is 5, which expression can be used to find the value of y when x is 11?
Leokris [45]

Answer: The value of y is 39.6 when x = 11 and the required expression is y(11) = 39.6

Step-by-step explanation:

Since we have given that

y varies directly as x.

If y = 18, when x = 5.

So, we need to find the value of y when x is 11.

According to question, we get that

\dfrac{y}{x}=\dfrac{18}{5}=\dfrac{y}{11}\\\\18\times 11=5y\\\\198=5y\\\\\dfrac{198}{5}=y\\\\y=39.6

Hence, the value of y is 39.6 when x = 11 and the required expression is y(11) = 39.6

5 0
3 years ago
How to find the area under the standard normal curve between z 1.5 and z 2.5?
alisha [4.7K]
It is possible to calculate mathematically the area under the normal curve between any two values of z.
However, tables/software have been developed to give the areas under the normal curve to the left of particular values of z.  The function is the probability of Z<z, or P(Z<z).
The area between two values z1 and z2 (where z2>z1) is therefore 
P(Z<z2)-P(Z<z1).

For example, to find the area between  z1=1.5, z2=2.5
is 
P(Z<2.5)-P(Z<1.5)
=0.99379-0.93319
=0.06060

(above values obtained by software, such as R)
For example, the value P(Z<2.5) can be calculated using
P(Z<2.5)=erf(2.5/sqrt(2))/2+1/2
where erf(x) is a mathematical function that does not have an explicit formula (calculated by summation of series, or tabulated).
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