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3 years ago

Please help please please help i have tried everything i could think of but i cant figure it out

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

6 0

Answer:

$\frac{\sqrt[4]{10xy^3}}{2y}$

Step-by-step explanation:

I am assuming you want me to rationalize the denominator.

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Clea attends music lessons every 2 weeks, dance class every 3 weeks, and has a dental appointment every 6 weeks. If she attends

Basile [38]

In 6 weeks.

For this problem, you need to determine the least common multiple of the numbers given which is 2, 3, and 6. You can calculate the LCM of multiple numbers by pairs.

The LCM of 2 and 3 is 6 because 6 is the smallest possible number that can be evenly divided by both 2 and 3.

Now you need the LCM of 6 and 6. Which is also 6, because 6 is the smallest number that can be evenly divided by 6.

So the LCM of 2, 3, and 6 is 6. So in 6 weeks, Clea will once again have a music less, attend dance class, and have a dental appointment, all during the same week.

7 0

4 years ago

Which of the following relations has a domain of {-6, 10)?

Alexeev081 [22]

Use this
Brownies P pp

4 0

3 years ago

Which of the following is not a congruence theorem or postulate A. ASA B. SSS C. SAS D. AAA

natka813 [3]

Answer:

D. AAA is not a congruence postulate

Step-by-step explanation:

6 0

3 years ago

What is the correct answer?

IgorLugansk [536]

Answer:

Step-by-step explanation:

i think

3 0

3 years ago

A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat

Marizza181 [45]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

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

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 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

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)

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

We assume that a prior estimation for p would be $\hat p =0.5$ since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

5 0

4 years ago