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2 years ago

15

The Greatest Show on Earth Circus Tour of 1890 hold the record for the most performers involved in a circus act. One hundred sev

enty-five animals and 263 persons took part. Would it be true to say that about 40 percent of the performers were animals?

1 answer:

Natasha_Volkova [10]2 years ago

3 0

Answer:

yes

Step-by-step explanation:

175 + 263 = 438

to get 40% do 40/100=0.4

so 438 x 0.4 = 175.2

175 is about 40% of 438

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Find the best approximation of the volume of the sphere.

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2 years ago

(6x10^2)/(3x10-5) help what is this in standard form?

Tasya [4]

<u>Answer: </u>

The standard form of $\frac{6 \times 10^{2}}{3 \times 10^{-5}}$ is 20,00,0000

<u>Solution: </u>

Given that $\frac{6 \times 10^{2}}{3 \times 10^{-5}}$ ---- eqn 1

To write $\frac{6 \times 10^{2}}{3 \times 10^{-5}}$ in standard form,

We know that $\bold{\frac{1}{a^{-m}} = a^{m}}$ .So $\frac{1}{10^{-5}}$ becomes $10^{5}$ .

Now eqn 1 becomes,

$= \frac{6 \times 10^{2}}{3} \times 10^{5}$ ----- eqn 2

We know that $\bold{a^{m} \times a^{n}=a^{m+n}}$ , so $10^{2} \times 10^{5} = 10^{7}$

Now eqn 2 becomes,

$= \frac{6}{3} \times 10^{7}$

$= 2 \times 10^{7}$ ---- eqn 3

Expanding $10^{7}$ :

Here 10 is the base term and 7 is the exponent value. So base term 10 is multiplied by itself 7 times.

$10^{7} = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$

Now eqn 3 becomes,

$= 2 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$

= 20,00,0000

Hence the standard form of $\frac{6 \times 10^{2}}{3 \times 10^{-5}}$ is 20,00,0000

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2 years ago

Margin of error: 0.009; confidence level: 99%; p and a unknown

mafiozo [28]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.009}{2.58})^2}=20544.44$

And rounded up we have that n=20545

Step-by-step explanation:

Assuming this question: Use the data to find minium sample size required to estimate population proportion. Margin of error: 0.009, confidence level: 99%, p and q are unknown.

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".

Solution to the problem

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

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 $ME =\pm 0.009$ 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)

Assuming an estimation of p as $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.009}{2.58})^2}=20544.44$

And rounded up we have that n=20545

7 0

3 years ago