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3 years ago

Octavian became the emperor of Rome in 27 B.C. He ruled until his death in A.D. 14. How long did he rule as emperor?

Mathematics

2 answers:

Archy [21]3 years ago

6 0

He ruled for 40 years (January, 27 B.C until August, AD 14)

Lilit [14]3 years ago

5 0

When counting B.C the years go backward. For example 25 B.C would become 24 B.C. on new years. When counting A.D we're counting forward. Year 1 becomes year 2 and so forth. So knowing that what do you think?

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Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago

I need help I don’t understand it

tiny-mole [99]

Shidddd lil buddy, try the 3rd one

4 0

3 years ago

Mademuasel [1]

Answer:

I only

Step-by-step explanaition:

The domain (x) can not repeat in order for it to be a function.

3 0

2 years ago

Read 2 more answers

A coin is flipped 40 times and it lands on head 16 timed based on the experimental probability how many heads would you protect

crimeas [40]

24 abcdefghijklmnopqrstuvwxyz

4 0

3 years ago

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 98% conf

adelina 88 [10]

Answer:

$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

So the answer for this case would be n=1337 rounded up to the nearest integer

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=1.1$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =0.07 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

So the answer for this case would be n=1337 rounded up to the nearest integer

3 0

3 years ago