What is the simplified form of the expression? 1/3(21m + 27)

You have inherited land that was purchased for $25,000 in 1970. The value of the land has appreciated by

Darina [25.2K]

Answer:

Future value of land = 25,000[1.04]³⁰

Future value of land = $81,084 (Approx.)

Step-by-step explanation:

Given:

Purchase cost of land in 1970 = $25,000

Application rate = 4% per year

Find:

Price of land in 2000

Computation:

Number of year = 2000 - 1970 = 30 year

Future value of land = [Purchase cost of land][1+Application rate]ⁿ

F = P[1+r]ⁿ

Future value of land = 25,000[1+4%]³⁰

Future value of land = 25,000[1+0.04]³⁰

Future value of land = 25,000[1.04]³⁰

Future value of land = 25,000[3.24339]

Future value of land = 81,084.75

Future value of land = $81,084 (Approx.)

5 0

3 years ago

Please help me i really need help please please help me please

Sergeeva-Olga [200]

3250

You will need to Multiply 10x325

4 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%