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crimeas [40]
3 years ago
13

(07.05 LC) How many solutions does the equation 4y + 7 = 5 + 2 + 4y have? One Two None Infinitely many

Mathematics
2 answers:
Elza [17]3 years ago
8 0

Answer: Infinitely many

Step-by-step explanation:

First simplify the right side by adding 5 and 2

now you have the equation 4y+7=7+4y

That is the exact same thing so any number works.

Reil [10]3 years ago
7 0

just to add to the great reply above.

\bf \begin{matrix} 4y \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}+7=5+2~~\begin{matrix} +4y \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}\implies 7=7

whenever you end up with something like 0 = 0, or 7 = 7, is a flag that both equations are exactly the same, in this case, the one on the right-hand-side is really the one one the left-hand-side <u>in disguise</u>.

So the graph of one, is the same as the graph of the other, or put in another words, let's say the graph the first one, the second one when graphed, will just be pancaked on top of the first one, and any point whatsoever on the second one, matches with the first one, and since both lines continue infinitely, then infinitely many solutions.

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Since the new Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

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