Let m∠COB = 50°30’, m∠AOB = 70° and m∠AOC = 20°30’. Could point C be in the interior of ∠AOB? Why?

Mathematics

1 answer:

AleksAgata [21]3 years ago

5 0

Answer:

Point C cannot be inside ∠AOB

Step-by-step explanation:

we know that

If point C is in the interior of ∠AOB

then

m∠AOB=m∠AOC+m∠COB -----> by addition theorem postulate

Verify

Substitute the values

Remember that

1 degree is equal to 60 minutes

$30'+30'=60'=1\°$

$70\°=50\°30'+20\°30'\\70\°\neq 71\°'$

therefore

Point C cannot be inside ∠AOB

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There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

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

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

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.