Estimate 2/3 ×40 how do you estimate

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

you estimate usally by the nearest 100. in this case its by 10 so your estimate answer is 30 because your final answer is 26. is the last number is greter than 5 you round up and lower than 5 you round down

Step-by-step explanation:

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Write the equation of this circle in standard form

Svetradugi [14.3K]

Answer:

Standard equation of this circle is (x +5)² + y² = 9.

Step-by-step explanation:

The equation of a circle with center at (h,k) and a radius of r is:

(x - h)² + (y - k)² = r²

Subtitute the variables with h= -5, k = 0, and r = 3

(x - (-5))² + (y - 0)² =3²

(x + 5)² + y² = 9

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8 0

3 years ago

Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is partic

o-na [289]

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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".

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

Part a

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 $ME =\pm 0.02$ 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)

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$ . And replacing into equation (b) the values from part a we got: