Whats the simple interest of $400 after 5 years at a 3% annual rate?

A study was done to determine the average number of homes that a homeowner owns in his or her lifetime. For the 50 homeowners su

ipn [44]

Answer:

The 95% confidence interval for the true average number of homes that a person owns in his or her lifetime is (4,6.2).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 50 - 1 = 49

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.0096

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.0096\frac{3.8}{\sqrt{50}} = 1.1$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 5.1 - 1.1 = 4

The upper end of the interval is the sample mean added to M. So it is 5.1 + 1.1 = 6.2.

The 95% confidence interval for the true average number of homes that a person owns in his or her lifetime is (4,6.2).

7 0

2 years ago

The question I need help on is 15.5z= -77.5 . I really don't get it please help:)

melamori03 [73]

The answer to this question is z=-5

6 0

3 years ago

Read 2 more answers

Solve this please show your work <3

Nookie1986 [14]

Given:

The expression is

$\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=$

To find:

The simplified form of the expression.

Solution:

We have,

$\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=$

The expression $\sqrt[3]{48}$ can be written as

$\sqrt[3]{48}=\sqrt[3]{8\cdot 6}$

$\sqrt[3]{48}=\sqrt[3]{8}\cdot \sqrt[3]{6}$ $[\because \sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}]$

$\sqrt[3]{48}=2\cdot \sqrt[3]{6}$

$\sqrt[3]{48}=2\sqrt[3]{6}$

Therefore, $\sqrt[3]{48}=\sqrt[3]{8\cdot 6}=2\sqrt[3]{6}$ .

5 0

3 years ago

Factoring , <br> a² +4a - 5

givi [52]

the factors of the term $a^2+4a-5$ are $(a-1)(a+5)$

Step-by-step explanation:

We need to find factors of the term: $a^2+4a-5$

For finding factors:

We need to break the middle term such that their product is equal to product of first and last term and their sum is equal to middle term

Breaking the middle term and finding factors:

$a^2+4a-5$

$=a^2+5a-a-5$

$=a(a+5)-1(a+5)$

$=(a-1)(a+5)$

So, the factors of the term $a^2+4a-5$ are $(a-1)(a+5)$

Keywords: Finding factors

Learn more about finding factors at:

#learnwithBrainly

6 0

2 years ago

What is 1.04 to the 1/12 power

LiRa [457]

1.04 ^ 1/12 = 0.086666666
or
1.04 ^ 1/12 = 13/150

8 0

3 years ago