I don't understand the answer to the problem

A tractor was purchased for $30,000 six years ago. what is the value of the tractor today if the depreciation rate was 2.5%?

MrRissso [65]

Initial cost = $30,000
Depreciation rate = 2.5%

Depreciation expense per year = 30,000*2.5/100 = $750
In six years,
Depreciation = 6*750 = $4,500

Value of the tractor = Initial cost - Depreciation = $30,000 - $4,500 = $25,500

6 0

3 years ago

Find the area of a rectangle with the length of 1 3/4 foot and a width of 1 1/2 foot.

Stolb23 [73]

To find the Area you need to use the formula L x W!! Anyways, You have to multiply the Length x the width. Your Length is 1 3/4 and your width is 1 1/2. So 1 3/4 x 1 1/2 is 2 5/8!!

4 0

3 years ago

Prove that: 5^31–5^29 is divisible by 100.

ANEK [815]

Answer:

See explanation

Step-by-step explanation:

Consider the expression

$5^{31}-5^{29}$

First, factor it:

$5^{31}-5^{29}=5^{29}\cdot(5^2-1)=5^{29}\cdot (25-1)=24\cdot 5^{29}$

Note that

$100=25\cdot 4$

Then

$5^{31}-5^{29}=24\cdot 5^{29}=6\cdot 4\cdot 25\cdot 5^{27}=6\cdot 100\cdot 5^{27}$

This shows that number 100 is a factor of the expression $5^{31}-5^{29}$ and, therefore, this expression is divisible by 100.

5 0

3 years ago

The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.

Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If $Z \leq -2$ or $Z \geq 2$ , the outcome X is considered unusual.

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 $s = \frac{\sigma}{\sqrt{n}}$ .