Simplify a + 2 + a + 2 + a + 2

A building society offers a rate of 11% per annum simple interest. Beth-Ann deposit 24000 in the society for 15 months. Calculat

Nostrana [21]

Answer:

The amount of money due to Beth-Ann at the end of this period is $27, 300.

Step-by-step explanation:

The amount at the end of a period, when simple interest is being charged is:

Amount (A) = Principal (P) + Interest (I)

Here interest is computed using the formula:

$I=P\times R\times T$

It is provided that:

P = 24,000

R = 11%

T = 15 months = 1.25 years

Compute the Interest as follows:

$I=P\times R\times T$

$=24000\times 0.11\times 1.25\\=3300$

The interest earned is, I = $3,300.

Compute the amount of money due to Beth-Ann at the end of this period as follows:

$A=P+I$

$=24000+3300\\=27300$

Thus, the amount of money due to Beth-Ann at the end of this period is $27, 300.

7 0

3 years ago

At the pet owner’s meeting, there are 20 people who own dog, and 35people who own cat and 9 people own goldfish. Some people own

Margaret [11]

Answer:

4

Step-by-step explanation:

Given that At the pet owner’s meeting, there are 20 people who own dog, and 35people who own cat and 9 people own goldfish. Some people own both catand dog and some people own both gold fish and cat. No dog owners owngoldfish. There are twice as many people who own both goldfish and catthan people who own both dog and cat.

Total no of people =52

The venn diagram attached gives a clear picture

assuming x to be owners of both dog and cat. Then 2x is for both goldfish and cats.

Totalling all mutually exclusive we get

64-3x=52

x=4

No of owners who own both cat and dog =4

7 0

3 years ago

What is an example of a country with this economic system?

Viktor [21]

Answer:In a free market economy, the law of supply and demand, rather than a central government, regulates production and labor. Companies sell goods and services at the highest price consumers are willing to pay, while workers earn the highest wages companies are willing to pay for their services. A purely capitalist economy is a free market economy; the profit motive drives all commerce and forces businesses to operate as efficiently as possible to avoid losing market share to competitors.

Step-by-step explanation:

4 0

3 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then