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4 years ago

I need help with this problem! Please Help t=-2.74 + sqrt(2.74^(2)-4-4.9(-10)) -: 2*-4.9

Mathematics

2 answers:

Mekhanik [1.2K]4 years ago

7 0

Answer:

no real solutions

see below

Step-by-step explanation:

It looks like a quadratic formula

t= -2.74 + sqrt(2.74^(2)-4*-4.9*(-10)) -: 2*-4.9

$t = \frac{-2.74 +\sqrt{2.74^2 - 4*(-4.9)*(-10)} }{2*(-4.9)}$

$t = \frac{-2.74 + \sqrt{7.5076 -196} }{-9.8} = \frac{-2.74 + \sqrt{-188.4924} }{-9.8}$

$t = \frac{-2.74 + 13.73i}{-9.8 }$

i = $\sqrt{-1}$

Ulleksa [173]4 years ago

3 0

Answer:

Step-by-step explanation:

t=(-2.74 + sqrt(2.74^(2)-4*-4.9*(-10))) / -2*-4.9

t=(-2.74 + sqrt(7.5076-4*+49))/9.8

t=(-2.74 + sqrt(7.5076+45))/9.8

t=(-2.74 + sqrt(52.5076))/9.8

t=(-2.74/9.8)+(sqrt(52.5076))/9.8

this is simplest form, next will be rounded answers

t=0.45981763305

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a) $ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

b) $ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

If we replace the values into equation (a) for 1983 we got:

$ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

Part b

Since is the same confidence level the z value it's the same.

If we replace the values into equation (a) for 2008 we got:

$ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

3 0

4 years ago

I need help with this problem! Please Help t=-2.74 + sqrt(2.74^(2)-4*-4.9*(-10)) -: 2*-4.9

I need help with this problem! Please Help t=-2.74 + sqrt(2.74^(2)-4-4.9(-10)) -: 2*-4.9