Polygons. Find the measure of x

the population of Watesville decreases at a 1.6% per year.If the population was 62,500 in 2014 what will it be in 2020

vichka [17]

The population in 2020 will be 57657.5 approximately

<u>Explanation:</u>

Given:

Population in 2014, P = 62,500

Rate of decrease, r = 1.6%

Population in 2020, Q = ?

Number of years between 2014 to 2020 = 5 years

Population in 2020 = $P(1 - \frac{r}{100} )^t$

Substituting the value we get:

$Q = 62500(1- \frac{1.6}{100})^5\\ \\Q = 62500 ( 0.92252) \\\\Q = 57657.46$

Therefore, the population in 2020 will be 57657.5 approximately

7 0

3 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.

garik1379 [7]

Answer:

The 95% confidence interval would be given by $0.281 \leq \mu_1 -\mu_2 \leq 0.529$

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".

$\bar X_1 =9.2$ represent the sample mean 1

$\bar X_2 =8.8$ represent the sample mean 2

n1=27 represent the sample 1 size

n2=30 represent the sample 2 size

$\sigma_1 =0.3$ population standard deviation for sample 1

$\sigma_2 =0.1$ population standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =9.2-8.8=0.4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$