X/9+2x/2=1/3 X = ? How to solve

What is the difference between population parameters and estimates in a sample?

Anna35 [415]

Answer:

Using that sample, you calculate the corresponding sample characteristic, which is used to summarize information about the unknown population characteristic. The population characteristic of interest is called a parameter and the corresponding sample characteristic is the sample statistic or parameter estimate.

Step-by-step explanation:

If you collect a sample and calculate the mean and standard deviation, these are sample statistics.

4 0

3 years ago

The principal of Angelo carrera saving account is $9855 it earns 6% interest compounded quarterly for two quarter

malfutka [58]

Answer:

$ 297.87 interest

Step-by-step explanation:

Calculate the amount and the compound interest.

Amount = PNR Where P is Principal ; N is the no of Terms and R is the rate of interest

Amount of 1 st quarter = $ 9,855 x 6 % x $\frac{1}{4}$ = $147.83

Compound interest is $ 9,855 + 147.83= $ 10,002.83

Amount of 2 nd quarter , 10,002.83 x 6% x $\frac{1}{4}$ = $150.04

Compound Interest is 10,002.83 + 150.04 = $10,152.87 amount

10,152.87 - 9,855= $297.87 interest

8 0

3 years ago

Assume that 1 euro is equivalent to 0.9718 U.S. dollars. If you want to exchange $3850 U.S. for euros, how much will you rece

lions [1.4K]

3741.43 is your answer. Just do $3850 x 0.9718 and you'll get your answer. I hope this helps :)

5 0

4 years ago

Read 2 more answers

suppose a study estimated the population mean for a variable of interest using a 99% confidence interval. If the width of the es

Dovator [93]

Answer:

Population standard deviation, $\sigma$ = 3683.063 .

Step-by-step explanation:

We are given that the width of the estimated confidence interval i.e. 99% is 600 and the sample size used in estimating the mean is 1000 which means ; n = 1000 and width = 600

We know that Width of confidence interval = 2 * Margin of error

Margin of error = $Z_\frac{\alpha}{2} * \frac{\sigma}{\sqrt{n} }$ = 2.5758 * $\frac{\sigma}{\sqrt{1000} }$ {because at 1% significance level