Ask question

Ask question

All categories

2 years ago

5

If you leave $2300 in an account earning 3% interest, compounded daily, how much money will be in the account after 5 years? (Ro

und your answer to two decimal places.)

1 answer:

Triss [41]2 years ago

3 0

Answer:

$2672.20

Step-by-step explanation:

2300(1+(0.03)/(365))^(5*365)

=$2672.20

You might be interested in

What is the mean absolute deviation of the data set

Ludmilka [50]

Answer:

Mean absolute deviation. The mean absolute deviation of a dataset is the average distance between each data point and the mean. It gives us an idea about the variability in a dataset

Step-by-step explanation:

Hope this helps

7 0

2 years ago

Read 2 more answers

Solve s/10=7 The solution is s=

Alexeev081 [22]

Answer:

s=70

t=30

r=32

:D!!

10×7=70

6×5=30

3 0

2 years ago

Jesse takes his dog and cat for their annual vet visit. Jesse's dog weighs 23 pounds. The vet tells him his cats weight is 5/8 a

kherson [118]

14.375 pounds because 23 divided by 8 is 2.875. 2.875 times 5 is 14.375.

7 0

2 years ago

Read 2 more answers

Mr Matthews i7 rise of 56 bricks in three and a half hours if you work the consistent rate how many bricks could you lay per hou

ohaa [14]

Answer: need more information

Step-by-step explanation:

7 0

2 years ago

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and

Ahat [919]

Answer:

(0.4062, 0.5098)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

For this problem, we have that:

356 dies were examined by an inspection probe and 163 of these passed the probe. This means that $n = 365$ and $\pi = \frac{163}{356} = 0.458$

Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.

So $\alpha$ = 0.05, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975[tex], so [tex]z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.458 - 1.96\sqrt{\frac{0.458*0.542}{356}} = 0.4062$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.458 + 1.96\sqrt{\frac{0.458*0.542}{356}} = 0.5098$

The correct answer is

(0.4062, 0.5098)

5 0

2 years ago

Read 2 more answers