Answer:
Step-by-step explanation:
If the employer matches up to 4% of Jaime's salary,
first thing is to find what that percentage adds up to;
Jaime's total salary = $54,000
4% of the total = 0.04*54,000 = $2,160 per year
If the employer matches a dollar for dollar, Jamie will have an additional $2,160 from his employer's contribution in 1 year
Next,
Find Jamie's contribution per month; 2,160/12 = $180
Therefore, to maximize his employer's contribution, he needs to contribute $180 per month
Answer:
fraction of the day was Priya lost
Step-by-step explanation:
As per the statement:
Priya spent 1/2 of the day hiking.
⇒She spent for hiking = 1/2 of the day.
It is also given that:
She was lost 5/6 of the time that she was hiking.
⇒![\text{She lost} = \frac{5}{6} \times \text{She was hiking}](https://tex.z-dn.net/?f=%5Ctext%7BShe%20lost%7D%20%3D%20%5Cfrac%7B5%7D%7B6%7D%20%5Ctimes%20%5Ctext%7BShe%20was%20hiking%7D)
Substitute the given values we have;
of the day
therefore,
fraction of the day was Priya lost
Answer:
3. 7680
4. 18630
Step-by-step explanation:
3. solution:
200 * 38.4 = 200 *38 + 200 * 0.4
= 7600 +80
=7680
4. solution:
900 * 20.7 = 900 *20 + 900 * 0.7
= 18000 + 630
= 18630
Answer:
![n = 191](https://tex.z-dn.net/?f=n%20%3D%20191)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
In this problem, we have that:
![M = 0.07, \pi = 0.42](https://tex.z-dn.net/?f=M%20%3D%200.07%2C%20%5Cpi%20%3D%200.42)
We have to find n
![0.07 = 1.96*\sqrt{\frac{0.42*0.58}{n}}](https://tex.z-dn.net/?f=0.07%20%3D%201.96%2A%5Csqrt%7B%5Cfrac%7B0.42%2A0.58%7D%7Bn%7D%7D)
![0.07\sqrt{n} = 1.96*\sqrt{0.42*0.58}](https://tex.z-dn.net/?f=0.07%5Csqrt%7Bn%7D%20%3D%201.96%2A%5Csqrt%7B0.42%2A0.58%7D)
![0.07\sqrt{n} = 0.9674](https://tex.z-dn.net/?f=0.07%5Csqrt%7Bn%7D%20%3D%200.9674)
![\sqrt{n} = \frac{0.9674}{0.07}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B0.9674%7D%7B0.07%7D)
![\sqrt{n} = 13.82](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%2013.82)
![\sqrt{n}^{2} = (13.82)^{2}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%5E%7B2%7D%20%3D%20%2813.82%29%5E%7B2%7D)
![n = 190.9](https://tex.z-dn.net/?f=n%20%3D%20190.9)
So, rounded to the nearest integer
![n = 191](https://tex.z-dn.net/?f=n%20%3D%20191)