Steve has x+500 marbles
richard has x marbles
if steve gives richard 300 marbles then
x+500-300 x+300
so x+200 and x+300
richard has more and by 100
<span>Well 68/80 as a percent is 85 % .I got that by dividing 68 by 80 and them multiplying by 100. 100-85= 15. So therefore the dress decreased in price by 15%. </span>
Answer:
13
Step-by-step explanation:
count A up to C
Answer:
A sample size of 79 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
![\pi = 0.21](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.21)
The margin of error is:
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?
A sample size of n is needed.
n is found when M = 0.09. So
![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)
![0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}](https://tex.z-dn.net/?f=0.09%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.21%2A0.79%7D%7Bn%7D%7D)
![0.09\sqrt{n} = 1.96\sqrt{0.21*0.79}](https://tex.z-dn.net/?f=0.09%5Csqrt%7Bn%7D%20%3D%201.96%5Csqrt%7B0.21%2A0.79%7D)
![\sqrt{n} = \frac{1.96\sqrt{0.21*0.79}}{0.09}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%5Csqrt%7B0.21%2A0.79%7D%7D%7B0.09%7D)
![(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.21*0.79}}{0.09})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%5Csqrt%7B0.21%2A0.79%7D%7D%7B0.09%7D%29%5E%7B2%7D)
![n = 78.68](https://tex.z-dn.net/?f=n%20%3D%2078.68)
Rounding up to the nearest whole number.
A sample size of 79 is needed.