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2 years ago

Please help me if you love your mom

Mathematics

1 answer:

kari74 [83]2 years ago

8 0

Answer:

10.2 cm²

Step-by-step explanation:

regular hexagon made up of six equilateral triangles

A = (2 × 1.7) /2 × 6 = 10.2

or by the side length

A = 3√3/2 * 2² = 10.3

You might be interested in

Rename 28 percent as a decimal and fraction in its simplest form.

VARVARA [1.3K]

28% is 28/100 as a fraction but the simplest form is 14/50 and it can be simplified to 7/25. the decimal is 0.28

fraction is 7/25
decimal is 0.28

7 0

2 years ago

Read 2 more answers

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species

kotegsom [21]

Answer:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

Step-by-step explanation:

The logistic equation is the following one:

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that $P(0) = 80, K = 2000$ .

The number of fish tripled in the first year. This means that $P(1) = 3P(0) = 3(80) = 240$ .

Using the equation for P(1), that is, P(t) when $t = 1$ , we find the value of r.

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

$240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}$

$280*(2000 + 80(e^{r} - 1)) = 160000e^{r}$

$280*(2000 + 80e^{r} - 80) = 160000e^{r}$

$280*(1920 + 80e^{r}) = 160000e^{r}$

$537600 + 22400e^{r} = 160000e^{r}$

$137600e^{r} = 537600$

$e^{r} = \frac{537600}{137600}$

$e^{r} = 3.91$

Applying ln to both sides.

$\ln{e^{r}} = \ln{3.91}$

$r = 1.36$

This means that the expression for the size of the population after t years is:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

4 0

2 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .