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3 years ago

5000 at 12% compound quarterly compared interest earned after 5 year's

Mathematics

1 answer:

hichkok12 [17]3 years ago

8 0

Answer:

A = P(1+r÷100)^n

where A is amount after some days

r is the rate

n is the number of years

p is the principle (the amount of money before the interest)

A = 5000 ( 1 + 12÷100)^5

A = 8811.70

A = 8812

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If the cost of a 2.5 meter cloth is $30.5. What will be the cost of 22 meters ?

alisha [4.7K]

Answer:

268.40

Step-by-step explanation:

We can write a ratio to solve

2.5 meters 22 meters

----------------- = --------------

30.5 dollars x dollars

Using cross products

2.5 * x = 30.5 * 22

2.5x =671

Divide each side by 2.5

2.5x / 2.5 = 671/2.5

x =268.4

8 0

2 years ago

A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.

Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean $(\mu)$ = 500 and a standard deviation $(\sigma)$ = 100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = $\dfrac{Y-\mu}{\sigma}$

Z-score = $\dfrac{625-500}{100}$

⇒ Z-score = $\dfrac{125}{100}$

⇒Z-score =1.25

Therefore , the required z-score = 1.25

6 0

3 years ago

N 1968, it cost $54,000 for a 30-second commercial during the Super Bowl. In 1973, the cost for a 30-second commercial was $88,0

madreJ [45]

In 5 years, the price increased about 30%.
I am not good at perentages, so correct me if I'm wrong. Also this is an estimation, not the exact percentage.

8 0

3 years ago

A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou

Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question, $x = 5$ and $\sigma = 2$ . The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$ .

To solve for $\mu$ , the mean of this distribution, the only thing that needs to be found is the value of $z$ . Since

The problem stated that $P(X \le 5) = 69.85\% = 0.6985$ . Hence, $P(Z \le z) = 0.6985$ .

The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$ . To estimate $z$ from $P(Z \le z) = 0.6985$ , it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$ , $z > 0$ . As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$ . Besides, $P(Z < 0) = P(Z \le 0) = 0.50$ . As a result: