All categories

3 years ago

What is 8.5 in a fraction with a denominator of 4 ? How do you get it ?

2 answers:

8 0

8.5 as a fraction is 8.5/1. To make the denominator 4, you must multiply both the top and bottom by 4 (4x8.5/1x4). This gives you your answer, which is 34/4. To double check this, you just divided 34 by 4. This equals 8.5, which is the original number.

tangare [24]3 years ago

6 0

8.5 is equal to 8.5/1 as a fraction. To write it in a fraction with a denominator of 4, we both multiply the numerator and the denominator by 4:

8.5/1
= (8.5*4)/ (1*4) (both multiply numerator and denominator by 4 does not change the result).
= 34/4.

The final answer is 34/4.

Hope this helps~

You might be interested in

What is the slope for -3 , -2.5

weqwewe [10]

Answer:

Is there only one coordinate point? is that all the information you got?

7 0

3 years ago

Kayla bought a necklace on sale for 35% off. What

Oliga [24]

Answer:

$39

Step-by-step explanation:

60 x 0.39 = 21

60 - 21 = 39

Kayla bought a necklace on sale for $39.

6 0

3 years ago

Y=x^2-2x-5;y=x^3-2x^2-5x-9

ehidna [41]

Are the directions asking you to solve these ?

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: