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3 years ago

What’s the value of 27.43? Write answer in a fraction

Mathematics

2 answers:

Nezavi [6.7K]3 years ago

4 0

Answer:

As an improper fraction: 27 $\frac{43}{100}$

As a mixed fraction: $\frac{2743}{100}$

Step-by-step explanation:

Any number could be expressed as a fraction 27 is $\frac{27}{1}$ and since 0.43 is forty-three hundredths we can express it as $\frac{43}{100}$ .

To find the answer as a mixed fraction we have to put the whole number in the beginning and the fraction in the end. 27 $\frac{43}{100}$

As an improper fraction we have to add 27 / 1 + 43/100 = $\frac{2743}{100}$

Alina [70]3 years ago

4 0

Answer:2743/100

Step-by-step explanation:

27.43=2743/100

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A village was founded four hundred years ago by a group of 20 people. In this village, the population triples every one hundred

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Answer:

Step-by-step explanation:

Treat this like compound interest: Use A = P(1 + r)^t.

Here, P is the initial population and A is 3 times that, or 3P. Since P = 20 people, 3P = 60 people,

and this population is reached after 100 years.

We need to determine r, substitute its value into the formula A = P(1 + r)^t, and then determine the population of the village after 400 years.

60 = 20(1 + r)^100

Simplifying, 3 = (1 + r)^100.

Taking the natural log of both sides,

ln 3 = 100 ln (1 + r), or

ln 3

ln (1 + r) = ---------------

100

= 1.0986 / 100 = 0.01986

We must solve this for r. Raising e to the power ln (1 + r), on the left side of an equation, and raising e to the power 0. 01986 on the right side, we get:

1 + r = 3, so r must = 2.

Now find the pop of the village today. Use the same equation: A = P (1+r)^t.

A = 20(1 +2)^4 (hundreds),

A = 20(3)^4, or

A = 81

The population after 400 years is 81.

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Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

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".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

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