What is 35\42 in simplest form

1 answer:

4 0

To put a fraction into simplest form, you need to divide both the numerator and the denominator by the GCF of both of these numbers.

35 and 42 have a GCF of 7.

So, if we divide
35/7=5
42/7=6

Therefore, 35/42 in simplest form is 5/6

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Add. –32.7+(–0.2)<br> a. –32.9<br> b. –32.5<br> c. 32.5<br> d. 32.9

yKpoI14uk [10]

The answer to your question is -32.9

5 0

3 years ago

The distribution of heights of young women aged 18 to 24 is approximately normal with a mean of 64.5 inches and a standard devia

Lena [83]

The 68 - 95 - 99.7 rule, gives the basis to solve this question.

It says that for a normal distribution 95% of the results are between the mean minus 2 standard deviations and the mean plus 2 standard deviations.

Here:
mean = 64.5 inches,
standard deviaton = 2.5 inches

mean - 2 standard deviations = 64.5 inches - 5 inches = 59.5 inches

mean + 2 standard deviations = 64.5 inches + 5 inches = 69.5 inches

Then, the answer is that 95% of women range approximately between 59.5 inches and 69.5 inches.

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3 years ago

The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor

nadya68 [22]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

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 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: