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3 years ago

Help asap with Arcs! 30 Points

Mathematics

1 answer:

alexira [117]3 years ago

6 0

The answer is 15

these chords are congruent because the radius of the circle are perpendicular to the them
so we can find EF by making an equation:
$3x + 5 = x + 35 \\ 3x - x = 35 - 5 \\ 2x = 30 \\ x = \frac{30}{2} = 15$

so EF is = 3(15)+5=50
good luck

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Explain how to write the number that is 0.02 less than 1.423

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3 years ago

PLEASE HELP. !!!!!!!!!!

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Answer:

the last one bottom right

Step-by-step explanation:

well, did it in my head, but the idea is to calculate it out

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The manager of a local nightclub has recently surveyed a random sample of 280 customers of the club. She would now like to deter

Lunna [17]

Answer:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

Step-by-step explanation:

We have the following info given:

$\bar X = 35.6$ represent the sampel mean for the age of customers

$\sigma = 5$ represent the population standard deviation

$n = 280$ represent the sample size selected

We want to test if the mean age of her customers is over 35 so then the system of hypothesis for this case are:

Null hypothesis: $\mu \leq 35$

Alternative hypothesis $\mu >35$

The statistic for this case is given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing the data given we got:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

We can calculate the p value since we are conducting a right tailed test like this:

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

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