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2 years ago

The distance between earth and moon is 384000000 m. this distance can be expressed as ____m

Mathematics

1 answer:

zalisa [80]2 years ago

5 0

This distance can be expressed as 3.84× $10^{8}$ m.

What is distance with example?

Distance is a scalar quantity.
The distance of an object can be defined as the complete path travelled by an object.

Given the distance between the earth and the moon is 384000km

Representing in exponential notation,

=384000×1000m [∵1km=1000m]

=384× $10^{6}$ m

=3.84× $10^{8}$ m

Learn more about distance

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<u>The correct question is -</u>

The distance between the Earth and the moon is approximately 384000 km. this distance can be expressed as ____m.

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$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

$(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778$

And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

Step-by-step explanation:

We know the following info:

$\bar X_1 = 21.9$ sample mean for group 1

$\bar X_2 = 19.8$ sample mean for group 2

$s_1 = 3.4$ sample standard deviation for group 1

$s_2 = 3.5$ sample standard deviation for group 2

$n_1 = 200$ sample size group 1

$n_2 = 200$ sample size group 2

We want to find a confidence interval for the difference of means and the correct formula to do this is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

Now we just need to find the critical value. The confidence level is 0.95 then the significance is $1-0.95 =0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df= n_1 +n_2 -2= 200+200-2= 398$

The critical value for this case would be : $t_{\alpha/2}=1.966$

And replacing into the confidence interval formula we got:

$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

$(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778$

And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

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