Principle (P) = $ 2,500

Rate (R) = 14%

Time (T) = 1 year

Simple Interest (SI) = $\frac{PRT}{100}$

SI = $\frac{2500 * 14 * 1}{100} = \frac{25 *14}{1} = 350$

Hence, the answer is $350.

3 0

3 years ago

The burning time of a very large candle is normally distributed with mean of 2500 hours and standard deviation of 20 hours. Find

dezoksy [38]

Answer: C.-1.5

Step-by-step explanation:

Given: The burning time of a very large candle is normally distributed with mean $(\mu)$ of 2500 hours and standard deviation $(\sigma)$ of 20 hours.

Let X be a random variable that represent the burning time of a very large candle.

Formula: $Z=\dfrac{X-\mu}{\sigma}$

For X = 2470

$Z=\dfrac{2470-2500}{20}\\\\\Rightarrow\ Z=\dfrac{-30}{20}\\\\\Rightarrow\ Z=\dfrac{-3}{2}\\\\\Rightarrow\ Z=-1.5$

So, the z-score they corresponds to a lifespan of 2470 hours. =-1.5

Hence, the correct option is C.-1.5.

7 0

2 years ago

1 121 12321 1234321.......56th​

1 121 12321 1234321.......56th