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3 years ago

The equation |ax + b| = c must have ?

Mathematics

2 answers:

NARA [144]3 years ago

8 0

Answer: Option c

Step-by-step explanation:

1. By definition |ax + b| = c can be written as:

-ax-b=c if ax+b is greater than zero (ax+b>0)

ax+b=c if ax+b is equal or greater than zero (ax+b $\geq$ 0)

This means that this function has two solutions.

2. Then, |ax + b| is always greater than zero. Therefore, if c is less than zero, then the equation has no solution.

3. Therefore, the function can have two solutions if c>0 and no solution is c<0.

Then, the function can have 0,1 or 2 solutions.

ArbitrLikvidat [17]3 years ago

4 0

Answer:

0, 1, or 2 solutions

Step-by-step explanation:

The equation |ax + b| = c

The equation have absolute value symbol

Absolute value always gives us the positive number.

For absolute value function , we need to consider two cases

positive and negative.

|x|=x for positive , and |-x|=x for negative case

For negative case we include negative sign

So |ax + b| = c can be written as 2 equations

(ax+b)=c (ax+b)=-c

So we will get maximum of 2 solutions

The equation |ax + b| = c must have 0, 1, or 2 solutions

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b) n=523 rounded up to the nearest integer

Step-by-step explanation:

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The confidence interval for the mean is given by the following formula:

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The degrees of freedom are df=n-1=48-1=47

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that $z_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

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$530+2.01\frac{70}{\sqrt{48}}=550.308$

So on this case the 95% confidence interval would be given by (509.592;550.308)

Part b

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Assuming that $\hat \sigma =s$

And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval is provided, $z_{\alpha/2}=1.96$ , replacing into formula (b) we got:

$n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523$

So the answer for this case would be n=523 rounded up to the nearest integer

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