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3 years ago

H e l p ? p l e a s e

Mathematics

1 answer:

Bingel [31]3 years ago

6 0

To clear the fractions we multiply both sides by the least common multiple of all the denominators.

1/2 x + 2/3 = 4

Denominators 2 and 3, so multiply both sides by [Answer]: 6

3/4 x + 1 = 5/6

Denoms 4 and 6, LCM=12 Answer: 12

6/7 x - 2/3 = 5/21

LCM(7,3,21)=21 Answer: 21

3/5 + x/2 = 9

LCM(5,2) Answer: 10

25/4 = 6 + 1/2 x

LCM(4,2) Answer: 4

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This is a dumb question but what is 68 time 100

Whitepunk [10]

68 x 100 = 6,800

Hope this helps

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2 years ago

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A 5-column table has 4 rows. The first column has entries A, B, C, Total. The second column is labeled X with entries 45, 20, 25

Setler [38]

Answer:

Step-by-step explanation:

To see if two events are independent you can use the formula P(a|b) = P(a)

In this problem P(C|Y) = 35/75 = .4667 and P(C) = 110/300 = 3.667,

therefore C and Y are not independent events because P(C∣Y) ≠ P(C).

The answer is choice D

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3 years ago

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You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$