What is the product of -13/8 and 24/5

The following information is known about a loan.

Katyanochek1 [597]

Use the formula I = PRT.....you'll get $2000

6 0

2 years ago

Read 2 more answers

Of 350 solitaire games that jay played, he won 154 times. what percent of the games did jay win?

ludmilkaskok [199]

Answer:

44.044%

Step-by-step explanation:

To find the answer we first have to find the value of each percent, to do this we divide 100 by 350, that gets us approximately 0.286. We know that he won 154 of them we then multiply 154 by 0.286 to get the percentage of games he won and we get 44.044.

5 0

3 years ago

Read 2 more answers

Elizabeth jogged the same distance for 2 days and she jogged 6 miles on the third day. She jogged for a total of 16 miles over 3

trasher [3.6K]

Answer: Elizabeth jogged 5 miles on the first day.

Step-by-step explanation:

16-6= 10

10/2 = 5

4 0

2 years ago

What percentage of the flag is taken up by the union?

svetoff [14.1K]

I don’t know really but my guess is 62%

3 0

2 years ago

According to a study conducted in one city, 39% of adults in the city have credit card debts of more than $2000. A simple random

Butoxors [25]

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.39, n = 100$

Then

$s = \sqrt{\frac{0.39*0.61}{100}} = 0.0488$

By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

5 0

2 years ago